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# Important Questions for CBSE Class 12 Physics Chapter 5 - Magnetism and Matter 2024-25

Last updated date: 15th Jul 2024
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## CBSE Class 12 Physics Chapter-5 Important Questions - Free PDF Download

The important questions for Class 12 Physics Chapter 5 PDF provide a deep insight into the questions that will help students in their exam preparation. The magnetism and matter class 12 important questions PDF cover all the important questions from chapter 5 of class 12 physics. The Important Questions Class 12 Physics Chapter 5 PDF will be highly beneficial for students who are preparing for the board examinations. The important questions for class 12 physics chapter 5 will boost the confidence of the students and enable them to tackle every type of question from chapter 5 of class 12 physics.

Also, check CBSE Class 12 Physics Important Questions for other chapters:

 CBSE Class 12 Physics Important Questions Sl.No Chapter No Chapter Name 1 Chapter 1 Electric Charges and Fields 2 Chapter 2 Electrostatic Potential and Capacitance 3 Chapter 3 Current Electricity 4 Chapter 4 Moving Charges and Magnetism 5 Chapter 5 Magnetism And Matter 6 Chapter 6 Electromagnetic Induction 7 Chapter 7 Alternating Current 8 Chapter 8 Electromagnetic Waves 9 Chapter 9 Ray Optics and Optical Instruments 10 Chapter 10 Wave Optics 11 Chapter 11 Dual Nature of Radiation and Matter 12 Chapter 12 Atoms 13 Chapter 13 Nuclei 14 Chapter 14 Semiconductor Electronic: Material, Devices And Simple Circuits 15 Chapter 15 Communication Systems
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## Study Important Questions for Class 12 Physics Chapter- 5 Magnetism and Matter

Very Short Answer Questions (1 Mark)

1. How does the intensity of magnetization of a paramagnetic material vary with increasing applied magnetic field?

Ans: The intensity of magnetization increases with the increase in applied magnetic field.

2. An iron bar magnet is heated to ${{1000}^{0}}C$ and then cooled in a magnetic field free space. Will it retain magnetism?

Ans: Curie temperature of iron is ${{770}^{0}}C$. When it is heated to a very high temperature magnetism of iron is lost and does not retain its magnetism further.

3. How will the magnetic field intensity at the centre of a circular wire carrying current change, if the current through the wire is doubled and radius of the coil is halved?

Ans: As $B\text{ }=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2\pi I}{r}$

$\Rightarrow B'\text{ }=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2\pi \left( 2I \right)}{r/2}$

$\Rightarrow B'\text{ }=4\left( \dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2\pi I}{r} \right)$

$\Rightarrow B'=4B$

4. Can neutrons be accelerated in a cyclotron? Why?

Ans:  No, neutrons cannot be accelerated in a cyclotron. This is because a neutron is neutral and a cyclotron can accelerate only charged particles.

5. What type of magnetic material is used in making permanent magnets?

Ans: Materials having high coercivity are used in making permanent magnets.

6. Which physical quantity has the unit $Wb/{{m}^{2}}$ ? Is it a scalar or a vector quantity?

Ans: Magnetic field has the unit $Wb/{{m}^{2}}$. It is a vector quantity.

1. A bar magnet of magnetic moment M is aligned parallel to the direction of a uniform magnetic field B. What is the work done to turn the magnet, so as the align its magnetic moment:

(i). Opposite to the field direction?

Ans: We know that work done, $W\text{ }=\text{ }MB\left( \cos {{\theta }_{1}}-\cos {{\theta }_{2}} \right)$

Here, ${{\theta }_{1}}={{\text{0}}^{\text{0}}}\text{ and }{{\text{ }\!\!\theta\!\!\text{ }}_{\text{2}}}\text{=18}{{0}^{0}}$

$\Rightarrow W\text{ }=\text{ }MB\left( \cos {{0}^{0}}-\cos {{180}^{0}} \right)$

$\Rightarrow W\text{ }=\text{ }MB\left( 1-\left( -1 \right) \right)$

$\Rightarrow W\text{ }=\text{ 2}MB$

(ii). Normal to the field direction?

Ans: Here, ${{\theta }_{1}}={{\text{0}}^{\text{0}}}\text{ and }{{\text{ }\!\!\theta\!\!\text{ }}_{\text{2}}}\text{=9}{{0}^{0}}$

$\Rightarrow W\text{ }=\text{ }MB\left( \cos {{0}^{0}}-\cos {{90}^{0}} \right)$

$\Rightarrow W\text{ }=\text{ }MB$

2. An electron in the ground state of hydrogen atom is revolving in anti - clock wise direction in a circular orbit. The atom is placed normal to the electron orbit makes an angle of ${{30}^{0}}$ in the magnetic field. Find the torque experienced by the orbiting electron?

Ans: In the above question it is given that:

Magnetic moment associated with electron $M\text{ }=\dfrac{eh}{4\pi {{m}_{e}}}$

$\theta ={{30}^{0}}$ and

$\tau =MB\sin \theta$

$\tau \text{ }=\dfrac{eh}{4\pi {{m}_{e}}}B\times \sin {{30}^{0}}\text{ }=\dfrac{eh}{8\pi {{m}_{e}}}$, which is the torque.

3. Define angle of dip. Deduce the relation connecting angle of dip and horizontal component of earth’s total magnetic field with the horizontal direction.

Ans: We know that:

$\dfrac{{{B}_{H}}}{B}=\cos \delta$ and

$\dfrac{{{B}_{V}}}{B}=\sin \delta$

$\Rightarrow \dfrac{\sin \delta }{\cos \delta }=\dfrac{{{B}_{V}}}{B}\times \dfrac{B}{{{B}_{H}}}$

$\Rightarrow \tan \delta =\dfrac{{{B}_{V}}}{{{B}_{H}}}$

4. A point change +q is moving with speed perpendicular to the magnetic field B as shown in the figure. What should be the magnitude and direction of the applied electric field so that the net force acting on the charge is zero?

Ans: We know that:

Force on the charge due to magnetic field $=\text{ }qVB\text{ }sin\theta$

Since $\overset{\to }{\mathop{B}}\,\bot$ to the plane of paper,

$F\text{ }=\text{ }qVB\text{ }sin{{90}^{0}}$

$F=\text{ }qVB\text{ }$ (along OY)

Force on the charge due to electric field is:

$F\text{ }=\text{ }qE$

Net force on charge is zero if $qE\text{ }=\text{ }qVB$

$E\text{ }=\text{ }VB$ (along YO)

5. The energy of a charged particle moving in a uniform magnetic field does not change. Why?

Ans: The force on a charged particle in a uniform magnetic field always acts in a direction perpendicular to the motion of the charge. As the work done by the magnetic field on the charge is zero, hence energy of the charged particle will not change.

6. In the figure, straight wire AB is fixed; white the loop is free to move under the influence of the electric currents flowing in them. In which direction does the loop begin to move? Justify.

Ans: As the current in AB and arm PQ are in the same direction therefore wire will attract the arm PQ with a force (say ${{F}_{1}}$ ). But repels the arm RS with a force (say ${{F}_{2}}$ ). Since arm PQ is closer to the wire AB, ${{F}_{1}}>{{F}_{2}}$ i.e., the loop will move towards the wire.

7. State two factors by which voltage sensitivity of a moving coil galvanometer can be increased.

Ans: We know that:

$Voltage\text{ }sensitivity\text{ }=\dfrac{nBA}{kr}$

It can be increased by

(1) increasing B using powerful magnets.

(2) decreasing k by using phosphor borne strips.

8. What is the magnetic moment associated with a coil of 1 turn, area of cross- section ${{10}^{-4}}{{m}^{2}}$ carrying a current of 2A?

Ans: We know that:

$m\text{ }=\text{ }NIA$

$\Rightarrow m\text{ }=\text{ 1}\times \text{1}{{\text{0}}^{4}}\times 2$

$\Rightarrow m\text{ }=\text{ 2}\times \text{1}{{\text{0}}^{4}}A{{m}^{2}}$

9. A Rowland ring of mean radius 15 cm has 3500 turns of wire wound on a ferromagnetic core of relative permeability 800. What is the magnetic field B in the core for a magnetising current of 1.2 A?

Ans: In the above question it is given that:

Mean radius of a Rowland ring, $r=\text{ }15\text{ }cm\text{ }=\text{ }0.15\text{ }m$

Number of turns on a ferromagnetic core, $N=\text{ }3500$

Relative permeability of the core material, ${{\mu }_{r}}=800$

Magnetising current, $I=\text{ }1.2\text{ }A$

The magnetic field is given by the relation:

$B=\dfrac{{{\mu }_{0}}{{\mu }_{r}}lN}{2\pi r}$

Where,

${{\mu }_{0}}=Permeability\text{ }of\text{ }free\text{ }space\text{ }=4\pi \times {{10}^{-7}}Tm{{A}^{-1}}$

$B=\dfrac{4\pi \times {{10}^{-7}}\times 800\times 1.2\times 3500}{2\pi \times 0.15}=4.48T$

Thus, the magnetic field in the core is $4.48T$.

10. At a certain location in Africa, a compass points $12{}^\circ$ west of the geographic north. The north tip of the magnetic needle of a dip circle placed in the plane of magnetic meridian points above the horizontal. The horizontal component of the earth's field is measured to be 0.16 G. Specify the direction and magnitude of the earth's field at the location.

Ans: In the above question it is given that:

Angle of declination, $\theta =12{}^\circ$

Angle of dip, $\delta ={{60}^{0}}$

Horizontal component of earth's magnetic field, ${{B}_{H}}=\text{ }0.16\text{ }G$

Earth's magnetic field at the given location $=\text{ }B$

We can relate B and ${{B}_{H}}$ as:

${{B}_{H}}=B\cos \delta$

$\Rightarrow B=\dfrac{{{B}_{H}}}{\cos \delta }$$=\dfrac{0.16}{\cos {{60}^{0}}}=0.32G$

Earth's magnetic field lies in the vertical plane, $12{}^\circ$ West of the geographic meridian, making an angle of $60{}^\circ$ (upward) with the horizontal direction. Its magnitude is $0.32G$.

11. A magnetic needle free to rotate in a vertical plane parallel to the magnetic meridian has its north tip pointing down at $22{}^\circ$with the horizontal. The horizontal component of the earth’s magnetic field at the place is known to be$0.35\,\,G$. Determine the magnitude of the earth’s magnetic field at the place.

Ans: It is provided that,

The horizontal component of earth’s magnetic field, ${{B}_{H}}=0.35G$

The angle made by the needle with the horizontal plane (angle of dip)$=\delta =22{}^\circ$.

Earth’s magnetic field strength is $B$.

We can relate $B$ and ${{B}_{H}}$ as: ${{B}_{H}}=B\cos \delta$

$\Rightarrow B=\dfrac{{{B}_{H}}}{\cos \delta }$

$\Rightarrow B=\dfrac{0.35}{\cos 22{}^\circ }=0.377\,G$

Clearly, the strength of earth’s magnetic field at the given location is$0.377\,G$.

12. If the solenoid in Exercise 5.5 is free to turn about the vertical direction and a uniform horizontal magnetic field of $0.25\,T$ is applied, what is the magnitude of torque on the solenoid when its axis makes an angle of $30{}^\circ$ with the direction of applied field?

Ans: Given is the magnetic field strength, $B=0.25\,T$

Magnetic moment, $M=0.6\,/T$

The angle, $\theta$ between the axis of the turns of the solenoid and the direction of the external applied field is $30{}^\circ$ .

Hence, the torque acting on the solenoid is given as:

$\tau =MB\sin (\theta )$

$\Rightarrow \tau =0.6\times 0.25\sin (30{}^\circ )$

$\Rightarrow \tau =7.5\times {{10}^{-2}}J$

Hence the magnitude of torque is $=7.5\times {{10}^{-2}}J$.

13. A closely wound solenoid of $800$turns and area of cross section $2.5\times {{10}^{-4}}\,{{m}^{2}}$carries a current of $3.0A$. Explain the sense in which the solenoid acts like a bar magnet. What is its associated magnetic moment?

Ans:  It is provided that the number of turns in the solenoid, $n=800$.

Area of cross-section, $A=2.5\times {{10}^{-4}}{{m}^{2}}$

Current in the solenoid, $I=3.0A$

A current-carrying solenoid is analogous to a bar magnet because a magnetic field develops along its axis, i.e., along its length joining the north and south poles.

The magnetic moment due to the given current-carrying solenoid is calculated as:

$M=nIA=800\times 3\times 2.5\times {{10}^{-4}}=0.6J/T$

Thus, the associated magnetic moment $=0.6J/T$

14. A short bar magnet placed with its axis at ${{30}^{0}}$ with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to $4.5\times {{10}^{-2}}J$. What is the magnitude of magnetic moment of the magnet?

Ans:  In the above question it is given that:

Magnetic field strength, $B=0.25T$

Torque on bar magnet, $T=4.5\times {{10}^{-2}}J$

Angle between the bar magnet external magnetic field, , $\theta ={{30}^{0}}$

We know that:

$T=MB\sin \theta$

$\Rightarrow M=\dfrac{T}{B\sin \theta }$

$\Rightarrow M=\dfrac{4.5\times {{10}^{-2}}}{0.25\sin {{30}^{0}}}=0.35J/T$

Hence, the magnetic moment is $0.35J/T$.

1. A short bar magnet of magnetic moment 0.9 J/T is placed with its axis at $\mathbf{6}{{\mathbf{0}}^{\mathbf{o}}}$ to a uniform magnetic field. It experiences a torque of 0.063 Nm.

(i). calculate the strength of the magnetic field

Ans: As $\tau =MB\sin \theta$,

$\theta ={{60}^{0}}$

$\tau =0.063\text{ }Nm$

$M=0.9\text{ }J/T$

$B=\dfrac{\tau }{M\sin \theta }=\dfrac{0.063}{0.9\sin {{60}^{0}}}=0.081T$

(ii). What orientation of the bar magnet corresponds to the equilibrium position in the magnetic field?

Ans: The magnet will be in stable equilibrium in the magnetic field if $\tau =0$

$\Rightarrow MB\sin {{0}^{0}}=0$ i.e., When the magnet aligns itself parallel to the field.

2. A beam of electrons is moving with a velocity of $3\times {{10}^{6}}m/s$ and carries a current of $1\text{ }\mu A$.

(a). How many electrons per second pass a given point?

Ans: We have $I=1\text{ }\mu A={{10}^{-6}}A$

And $n=\dfrac{I}{q}=\dfrac{{{10}^{-6}}}{1.6\times {{10}^{-19}}}=6.25\times {{10}^{12}}$

(b). How many electrons are in 1m of the beam?

Ans: We know that electrons traverse a distance of $3\times {{10}^{6}}m$per second.

Thus, number of electrons in one meter of the beam$=\dfrac{6.25\times {{10}^{12}}}{3\times {{10}^{6}}}=2.08\times {{10}^{6}}{{m}^{-1}}$

(c). What is the total force on all the electrons in 1m of the beam if it passes through the field of $0.1N{{A}^{-1}}{{m}^{-1}}$?

Ans: Force on one meter of the beam of electrons will be:

$F=\dfrac{0.1}{{{10}^{-6}}}={{10}^{5}}N$

3. What is the main function of soft iron core used in a moving coil galvanometer? A galvanometer gives full deflection for Ig. Can it be converted into an ammeter of range $I\text{ }<\text{ }Ig$?

Ans: Soft iron core is used in the moving coil galvanometer because it increases the strength of the magnetic field thus increases the sensitivity of the galvanometer.

We know that, $S\text{ }=\dfrac{G{{I}_{g}}}{I-{{I}_{g}}}$

For $I\text{ }<\text{ }Ig$, S becomes negative. Clearly, it cannot be converted into an ammeter of range I < Ig.

4. Two wires loops PQRSP formed by joining two semi-circular wires of different radii carry a current I as shown in the figure. What is the direction of the magnetic induction at the centre C.?

Ans: Magnetic field due to semicircle QR at C is

${{B}_{1}}=\dfrac{1}{2}\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2\pi I}{{{R}_{1}}}$

Magnetic field due to semicircle is at C is

${{B}_{2}}=\dfrac{1}{2}\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2\pi I}{{{R}_{2}}}$

Net field,

$B={{B}_{1}}-{{B}_{2}}$

$\Rightarrow B=\dfrac{2\pi I}{2}\dfrac{{{\mu }_{0}}}{4\pi }\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$

$\Rightarrow B=\dfrac{{{\mu }_{0}}I}{4}\left( \dfrac{1}{{{R}_{1}}}-\dfrac{1}{{{R}_{2}}} \right)$

5. A circular coil is placed in uniform magnetic field of strength 0.10T normal to the plane of coil. If current in the coil is 5.0A. Find.

(a) Total torque on the coil

Ans: We have:

$B\text{ }=\text{ }0.10T$

$\theta ={{0}^{0}}$(Normal to plane of the coil)

$I\text{ }=\text{ }5.0\text{ }A$,$Area\text{ }={{10}^{-5}}{{m}^{2}}$, $n={{10}^{29}}/{{m}^{3}}$

$\tau =MB\sin \theta =MB\sin {{0}^{0}}=0$, which is the required torque.

(b) Total force on the coil

Ans: Total force on the coil $=0N$

(c). Average force on each electron due to magnetic field (The coil is made of copper wire of cross- sectional area and free electron density in copper is ${{10}^{29}}/{{m}^{3}}$)

Ans: We know that:

${{F}_{av}}=q\left( {{{\vec{v}}}_{d}}\times \vec{B} \right)$

$\left( I\text{ }=\text{ }neA{{v}_{d}} \right)$

$\Rightarrow {{F}_{av}}=q\left( \dfrac{qI}{neA}\times B \right)$

$\Rightarrow {{F}_{av}}=\dfrac{IB}{nA}=\dfrac{5\times 0.10}{{{10}^{29}}\times {{10}^{-5}}}=5\times {{10}^{-25}}N$

6. Using Ampere’s circuital law, derive an expression for magnetic field along the axis of a Toroidal solenoid.

Ans: If n is the number of turns per unit length; I be the current flowing through the Toroid;

Then from Ampere’s circuital law

$\oint{\vec{B}.d\vec{l}={{\mu }_{0}}\times tot\text{al current flowing through tor}oid}$

$\Rightarrow \oint{\vec{B}.d\vec{l}={{\mu }_{0}}\left( 2\pi rnI \right)}$

$\Rightarrow \int\limits_{0}^{2\pi r}{Bdl\cos {{0}^{0}}={{\mu }_{0}}\left( 2\pi rnl \right)}$

$\Rightarrow B\int\limits_{0}^{2\pi r}{dl}={{\mu }_{0}}\left( 2\pi rnI \right)$

$\Rightarrow B.2\pi r={{\mu }_{0}}\left( 2\pi rnI \right)$

$\Rightarrow B={{\mu }_{0}}nI$, which is the required magnetic field here.

7. A short bar magnet of magnetic moment $M=0.32\,J/T$ is placed in a uniform magnetic field of $0.15\,T$. If the bar is free to rotate in the plane of the field, which orientation and would correspond to its

(a) Stable equilibrium? What is the potential energy of the magnet in this case?

Ans: It is provided that moment of the bar magnet, $M=0.32J/T$.

External magnetic field, $B=0.15T$

It is considered as being in stable equilibrium, when the bar magnet is aligned along the magnetic field. Therefore, the angle $\theta$, between the bar magnet and the magnetic field is $0{}^\circ$ .

Potential energy of the system $=-MB\cos (\theta )$

$\Rightarrow -MB\cos (\theta )=-0.32\times 0.15\times \cos (0)=-4.8\times {{10}^{-2}}J$

Hence the potential energy is $=-4.8\times {{10}^{-2}}J$.

(b) Unstable equilibrium? What is the potential energy of the magnet in this case?

Ans: It is provided that moment of the bar magnet, $M=0.32J/T$

External magnetic field, $B=0.15T$

When the bar magnet is aligned opposite to the magnetic field, it is considered as being in unstable equilibrium, $\theta =180{}^\circ$

Potential energy of the system is hence$=-MB\cos (\theta )$

$\Rightarrow -MB\cos (\theta )=-0.32\times 0.15\times \cos (180{}^\circ )=4.8\times {{10}^{-2}}J$

Hence the potential energy is $=4.8\times {{10}^{-2}}J$.

8. A closely wound solenoid of $2000$ turns and area of cross-section $1.6\times {{10}^{-4}}{{m}^{2}}$, carrying a current of $4.0\,A$, is suspended through its centre allowing it to turn in a horizontal plane.

(a) What is the magnetic moment associated with the solenoid?

Ans: Given is the number of turns on the solenoid, $n=2000$

Area of cross-section of the solenoid, $A=1.6\times {{10}^{-4}}{{m}^{2}}$

Current in the solenoid, $I=4A$

The magnetic moment inside the solenoid at the axis is calculated as:

$M=nAI=2000\times 1.6\times {{10}^{-4}}\times 4=1.28A{{m}^{2}}$

(b) What is the force and torque on the solenoid if a uniform horizontal magnetic field of $7.5\times {{10}^{-2}}T$ is set up at an angle of $30{}^\circ$ with the axis of the solenoid?

Ans: Provided that,

Magnetic field, $B=7.5\times {{10}^{-2}}T$

Angle between the axis and the magnetic field of the solenoid, $\theta =30{}^\circ$

Torque, $\tau =MB\sin (\theta )$

$\Rightarrow \tau =1.28\times 7.5\times {{10}^{-2}}\sin (30{}^\circ )$

$\Rightarrow \tau =4.8\times {{10}^{-2}}Nm$

Given the magnetic field is uniform, and the force on the solenoid is zero. The torque on the solenoid is $4.8\times {{10}^{-2}}Nm$.

9. A circular coil of $16$turns and radius $10\,cm$carrying a current of $0.75$A rests with its plane normal to an external field of magnitude $5.0\times {{10}^{-2}}\,T$. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of $2.0/s$. What is the moment of inertia of the coil about its axis of rotation?

Ans: It is provided that,

The number of turns in the given circular coil solenoid, $N=16$

Radius of the coil, $r=10cm=0.1m$

Cross-section of the coil, $A=\pi {{r}^{2}}=\pi \times {{(0.1)}^{2}}{{m}^{2}}$

Current in the coil, $I=0.75A$

Magnetic field strength, $B=5.0\times {{10}^{-2}}T$

Frequency of oscillations of the coil, $\upsilon =2.0/\operatorname{s}$

Therefore, magnetic moment, $M=NAI=NI\pi {{r}^{2}}$

$\Rightarrow M=16\times 7.5\times \pi \times {{0.1}^{2}}$

$\Rightarrow M=0.3777J/T$

Frequency is given by the relation:

$\nu =\dfrac{1}{2\pi }\sqrt{\dfrac{MB}{I}}$

where,

$I=$Moment of inertia of the coil

$\Rightarrow I=\dfrac{MB}{4{{\pi }^{2}}{{\nu }^{2}}}$

$\Rightarrow I=\dfrac{0.377\times 5\times {{10}^{-2}}}{4{{\pi }^{2}}\times {{2}^{2}}}$

$\Rightarrow I=1.19\times {{10}^{-4}}kg\,{{m}^{2}}$

Clearly, the moment of inertia of the coil about its axis of rotation $1.19\times {{10}^{-4}}kg\,{{m}^{2}}$.

10. A short bar magnet has a magnetic moment of $0.48\,J/T$. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of $10\,cm$from the centre of the magnet on

(a) The axis,

Ans: Provided that the magnetic moment of the given bar magnet, $M$is $0.48J/T$

Given distance, $d=10cm=0.1m$

The magnetic field at d-distance, from the centre of the magnet on the axis is given by the relation:

$B=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2M}{{{d}^{3}}}$

here,

${{\mu }_{0}}=$ Permeability of free space$=4\pi \times {{10}^{-7}}Tm/A$

Substituting these values, $B$ becomes as follows:

$\Rightarrow B=\dfrac{4\pi \times {{10}^{-7}}}{4\pi }\dfrac{2\times 0.48}{{{0.1}^{3}}}$

$\Rightarrow B=0.96\times {{10}^{-4}}T=0.96\,G$

The magnetic field is $0.96G$ along the South-North direction.

(b). The equatorial lines (normal bisector) of the magnet.

Ans: The magnetic field at a point which is $d=10cm=0.1m$ away on the equatorial of the magnet is given as:

$B=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{M}{{{d}^{3}}}$

$\Rightarrow B=\dfrac{4\pi \times {{10}^{-7}}}{4\pi }\dfrac{0.48}{{{0.1}^{3}}}$

$\Rightarrow B=0.48\times {{10}^{-4}}T=0.48G$

The magnetic field is $0.48G$along the North-South direction.

11. A short bar magnet placed in a horizontal plane has its axis aligned along the magnetic north-south direction. Null points are found on the axis of the magnet at $14\,cm$ from the centre of the magnet. The earth’s magnetic field at the place is $0.36\,\,G$and the angle of dip is zero. What is the total magnetic field on the normal bisector of the magnet at the same distance as the null–point (i.e., $14\,\,cm$) from the centre of the magnet? (At null points, field due to a magnet is equal and opposite to the horizontal component of earth’s magnetic field).

Ans: Provided that,

The magnetic field of Earth at the given place,$H=0.36\,G$

The magnetic field at a $d$-distance, on the axis of the magnet is given as:

${{B}_{1}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2M}{{{d}^{3}}}=H$

Here,

${{\mu }_{0}}=$ Permeability of free space$=4\pi \times {{10}^{-7}}Tm/A$

$M=$ The magnetic moment

The magnetic field at the same distance $d$, on the equatorial line of the magnet is given as:

${{B}_{2}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{M}{{{d}^{3}}}$

$\Rightarrow {{B}_{2}}=H/2$ (comparing with ${{B}_{1}}$)

Therefore, the total magnetic field,

$B={{B}_{1}}+{{B}_{2}}$

$\Rightarrow B=H+H/2$

$\Rightarrow B=0.36+0.18=0.54$

Clearly, the magnetic field is 0.54 G along the direction of earth’s magnetic field.

12. A long straight horizontal cable carries a current of $2.5\,A$ in the direction $10{}^\circ$ south of west to $10{}^\circ$ north of east. The magnetic meridian of the place happens to be $10{}^\circ$ west of the geographic meridian. The earth’s magnetic field at the location is $0.33\,G$, and the angle of dip is zero. Locate the line of neutral points (ignore the thickness of the cable). (At neutral points, magnetic field due to a current-carrying cable is equal and opposite to the horizontal component of earth’s magnetic field.)

Ans: Provided that,

Current in the wire, $I=2.5A$

The angle of dip at the location, $\delta =0{}^\circ$

The Earth’s magnetic field, $H=0.33G=0.33\times {{10}^{-4}}T$

The horizontal component of earth’s magnetic field is given as:

${{H}_{H}}=H\cos \delta =0.33\times {{10}^{-4}}\times \cos 0{}^\circ =0.33\times {{10}^{-4}}T$

The magnetic field at the neutral point at a distance R from the cable is given by the relation:

${{H}_{H}}=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{I}{R}$

here,  ${{\mu }_{0}}$ = Permeability of free space $=4\pi \times {{10}^{-7}}Tm/A$

$\Rightarrow R=\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{I}{{{H}_{H}}}$

$\Rightarrow R=\dfrac{4\pi \times {{10}^{-7}}}{2\pi }\dfrac{2.5}{0.33\times {{10}^{-4}}}=15.15\times {{10}^{-3}}m=1.51cm$

Clearly, a set of neutral points lie on a straight line parallel to the cable at a perpendicular distance of $1.51cm$ above the plane of the paper.

13. A compass needle free to turn in a horizontal plane is placed at the centre of circular coil of 30 turns and radius 12 cm. The coil is in a vertical plane making an angle of $45{}^\circ$ with the magnetic meridian. When the current in the coil is$0.35\,A$, the needle points west to east.

(a) Determine the horizontal component of the earth’s magnetic field at the location.

Ans: Provided that,

The number of turns in the given circular coil, $N=30$

The radius of the given circular coil, $r=12cm=0.12m$

Current in the coil, $I=0.35A$

Angle of dip, $\delta =45{}^\circ$

The magnetic field due to the current I, at a distance r, is given as:

${{B}_{{}}}=4\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{I}{r}$

here,

${{\mu }_{0}}$ = Permeability of free space $=4\pi \times {{10}^{-7}}Tm/A$

$\Rightarrow B=\dfrac{4\pi \times {{10}^{-7}}}{4\pi }\dfrac{2\pi \times 30\times 0.35}{0.12}$

$\Rightarrow B=54.9\times {{10}^{-4}}T=0.549G$

The compass needle points West to East. Hence, the horizontal component of earth’s magnetic field is given as:

${{B}_{H}}=B.\sin \delta$

$\Rightarrow {{B}_{H}}=0.549\,\sin 45{}^\circ =0.388G$

(b) The current in the coil is reversed, and the coil is rotated about its vertical axis by an angle of $90{}^\circ$ in the anticlockwise sense looking from above. Predict the direction of the needle. Take the magnetic declination at the places to be zero.

Ans: If the direction of the current flowing in the coil is reversed and if the coil is also rotated about its vertical axis by an angle of $90{}^\circ$, the needle will rearrange and reverse its original direction. In the given case, the needle would point from East to West.

14. A magnetic dipole is under the influence of two magnetic fields. The angle between the field directions is $60{}^\circ$, and one of the fields has a magnitude of $1.2\times {{10}^{-2}}T$. If the dipole comes to stable equilibrium at an angle of $15{}^\circ$with this field, what is the magnitude of the other field?

Ans: Provided that,

Magnitude of one of the magnetic fields, ${{B}_{1}}=1.2\times {{10}^{-2}}T$

Magnitude of the other magnetic field is ${{B}_{2}}$.

Angle between the above-mentioned two fields, $\theta =60{}^\circ$

At the state of stable equilibrium, the angle between the dipole and field${{B}_{1}}$ is ${{\theta }_{1}}=15{}^\circ$

Angle between the dipole and field ${{B}_{2}}$ is ${{\theta }_{2}}=\theta -{{\theta }_{1}}=60{}^\circ -15{}^\circ =45{}^\circ$

At a rotational equilibrium, the torques experienced by the dipole, due to both the fields, must balance each other.

Therefore, torque due to field ${{B}_{1}}\,\,=$Torque due to field ${{B}_{2}}$

$M{{B}_{1}}\sin {{\theta }_{1}}=M{{B}_{2}}\sin {{\theta }_{2}}$

Where,

$M=$Magnetic moment of the dipole

$\Rightarrow {{B}_{2}}=\dfrac{{{B}_{1}}\sin {{\theta }_{1}}}{\sin {{\theta }_{2}}}$

$\Rightarrow {{B}_{2}}=4.39\times {{10}^{-3}}T$

Clearly, the magnitude of the other magnetic field is $4.39\times {{10}^{-3}}T$.

15. The magnetic moment vectors ${{\mu }_{s}}$and ${{\mu }_{l}}$associated with the intrinsic spin angular momentum $\vec{S}$ and orbital angular momentum $\vec{l}$, respectively, of an electron are predicted by quantum theory (and verified experimentally to a high accuracy) to be given by

${{\mu }_{s}}=-(e/m)S$

${{\mu }_{l}}=-(e/2m)l$

Which of these relations is in accordance with the result expected classically? Outline the derivation of the classical result.

Ans: According to the definition of magnetic moment-${{\mu }_{l}}$ and orbital angular momentum-$l$.

Magnetic moment associated with the motion of the electron is:

${{\mu }_{l}}=iA=-(e/T).\pi {{r}^{2}}$

And the corresponding angular momentum is:

$l=mvr=m(2\pi r/T){{r}^{{}}}$

Where $r$ is the radius of the orbit, which the mass of electron is $m$and its charge $(-e)$ completes in time $T$.

Dividing ${{\mu }_{l}}\,\,by\,\,l$, one would get:

$\dfrac{{{\mu }_{l}}}{l}=\dfrac{-e}{T}.\pi {{r}^{2}}\times \dfrac{T}{{{m}^{2}}\pi {{r}^{2}}}=-\dfrac{e}{2m}$

$\Rightarrow {{\vec{\mu }}_{l}}=-\left( \dfrac{e}{2m} \right)\vec{l}$

Evidently, it can be seen that ${{\vec{\mu }}_{l}}$ and $\vec{l}$ will be antiparallel (both being normal to the plane of the orbit).

In contrast, $\dfrac{{{\mu }_{s}}}{s}=\left( \dfrac{e}{m} \right)$ and it is derived on the basis of quantum mechanics and is verified experimentally.

1. A particle of mass m and charge q moving with a uniform speed normal to a uniform magnetic field B describes a circular path of radius & Derive expressions for

(1) Radius of the circular path (2) time period of revolution (3) Kinetic energy of the particle?

Ans: A particle of mass (m) and change (q) moving with velocity normal to describes a circular path if

$\dfrac{m{{v}^{2}}}{r}=qBv\sin \theta$

$\Rightarrow \dfrac{m{{v}^{2}}}{r}=qBv\left( \because \theta =90{}^\circ \right)$

$\Rightarrow r=\dfrac{mv}{Bq}$ …… (1)

This is the required radius of the circular path.

Now, since

Time period of Revolution $during\text{ }circular\text{ }path\text{ }=\dfrac{Circumf\text{erence of c}ircle}{velocity}$;

$T=\dfrac{2\pi r}{v}$

$\Rightarrow T=\dfrac{2\pi r.m}{Bqr}$ ……. (from 1)

$\Rightarrow T=\dfrac{2\pi m}{Bq}$ …… (2)

This is the required time period.

Now,

$\text{Kinetic energy K}\text{.E}=\dfrac{1}{2}m{{v}^{2}}$

$\Rightarrow \text{K}\text{.E}=\dfrac{1}{2}m{{\left( \dfrac{Bqr}{m} \right)}^{2}}$

$\Rightarrow \text{K}\text{.E}=\dfrac{{{B}^{2}}{{q}^{2}}{{r}^{2}}}{2m}$, which is the required kinetic energy.

2. Write an expression for the force experienced by the charged particle moving in a uniform magnetic field B with the help of labeled diagram explain the working of cyclotron. Show that cyclotron frequency does not depend upon the speed of the particle.

Ans: Force experienced by the charged particle moving at right angles to uniform magnetic field $\vec{B}$with velocity $\vec{v}$ is given by $F=\text{ }q\text{ }\left( \text{\vec{v}}\times \text{\vec{B}} \right)$ Initially Dee ${{D}_{1}}$ is negatively charged and Dee ${{D}_{2}}$ is positively charged so, the positive ion will get accelerated towards Dee ${{D}_{1}}$. Since the magnetic field is uniform and acting at right angles to the plane of the Dees so the ion completes a circular path in ${{D}_{1}}$when ions come out into the gap, polarity of the Dee’s gets reversed used the ion is further accelerated towards Dee ${{D}_{2}}$ with greater speed and cover a bigger semi-circular path. This process is separated time and again and the speed of the ion becomes faster till it reaches the periphery of the dees where it is brought out by means of a deflecting plate and is made to bombard the target.

Since $F\text{ }=\text{ }qVBsin{{90}^{0}}$ provides the necessary centripetal force to the ion to cover a circular path so we can say $\dfrac{m{{v}^{2}}}{r}=qvB$

$r=\dfrac{mv}{Bq}$ …… (1)

Time period = $\dfrac{2\pi r}{v}=\dfrac{2\pi rm}{Bqr}=\dfrac{2\pi m}{Bq}$

$V=\dfrac{1}{T}=\dfrac{Bq}{2\pi m}$

Thus, frequency is independent of velocity.

3.

(a) Obtain an expression for the torque acting on a current carrying circular loop.

Ans: ABCD is a square loop of length (L) and area (A). Let I be the current flowing in the anticlockwise direction. Let $\theta$ be the angle between the normal to the loop and magnetic field $\vec{B}$.

Force acting on arm AB of the loop

${{\vec{F}}_{1}}=I\left( \vec{L}\times \vec{B} \right)\left( outwards \right)$

Force on arm CD

${{\vec{F}}_{2}}=I\left( \vec{L}\times \vec{B} \right)\left( inwards \right)$

Force on arm BC

${{\vec{F}}_{3}}=I\left( \vec{L}\times \vec{B} \right)\left( downwards \right)$

Force on arm DA

${{\vec{F}}_{4}}=I\left( \vec{L}\times \vec{B} \right)\left( upwards \right)$

Since ${{\vec{F}}_{3}}$ and ${{\vec{F}}_{4}}$ are equal and opposite and also acts along the same line, hence they cancel each other.

${{\vec{F}}_{1}}$ and ${{\vec{F}}_{2}}$ are also equal and opposite but their line of action is different, so they form a couple and makes the rectangular loop rotate anti clockwise.

Thus $\tau =\text{ }either\text{ }force\times \text{ }\bot distance$

$\tau =I\left( \vec{L}\times \vec{B} \right)\times DN$

$\tau =I\left( \vec{L}\times \vec{B} \right)\times b\sin \theta$

$\tau =ILB\sin {{90}^{0}}\times b\sin \theta$

$\tau =IAB\sin \theta$

For loop of N turns

$\tau =NIAB\sin \theta$

$\tau =MB\sin \theta \left( \because M=NIA \right)$

$\vec{\tau }=\vec{M}\times \vec{B}$

Where M is magnetic moment of the loop.

(b). What is the maximum torque on a galvanometer coil 5 cm $\times$ 12 cm of 600 turns when carrying a current of  $\mathbf{1}{{\mathbf{0}}^{-\mathbf{5}}}$ A in a field where flux density is $0.10Wb/{{m}^{2}}$?

Ans: It is known that

$\tau =NIAB\sin \theta$

Torque will be maximum when $\theta =\text{ }{{90}^{o}}$

${{\tau }_{\max }}=NIAB\left( \because \sin {{90}^{0}}=1 \right)$

${{\tau }_{\max }}=600\times {{10}^{-5}}\times \left( 0.10 \right)\left( 60\times {{10}^{-4}} \right)=3.6\times {{10}^{-6}}Nm$.

4. The current sensitivity of a moving coil galvanometer increases by 20% when its resistance is increased by a factor of two. Calculate by what factor, the voltage sensitivity changes?

Ans: We know that:

Current sensitivity, $\dfrac{\alpha }{I}=\dfrac{nBA}{k}$ …… (i)

Voltage sensitivity, $\dfrac{\alpha }{V}=\dfrac{nBA}{kr}$ …… (ii)

Resistance of a galvanometer increases when n and A are changed

Given $R'\text{ }=\text{ }2R$

Then $n\text{ }=n'$ and $A\text{ }=A'$

New current sensitivity

$\dfrac{\alpha '}{I'}=\dfrac{n'BA'}{k}$ …… (iii)

New voltage sensitivity

$\dfrac{\alpha '}{V}=\dfrac{\alpha '}{I'R'}=\dfrac{n'BA'}{2kr}$ …… (iv)

Since, $\dfrac{\alpha '}{I'}=\dfrac{120\alpha }{100I}$ …… (v)

From (i) and (iii)

$\dfrac{n'A'B}{R}=\dfrac{\alpha 120}{I100}$

$\dfrac{n'A'B}{R}=\dfrac{nAB120}{k100}$

$n'A'=\dfrac{6}{5}nA$

Using equation (iv)

$\dfrac{\alpha '}{V}=\dfrac{6}{5}\dfrac{nBA}{2kr}=\dfrac{3}{5}\dfrac{\alpha }{V}$

Thus, voltage sensitivity decreases by a factor of $\dfrac{3}{5}$.

5.

(a) Show how a moving coil galvanometer can be converted into an ammeter.

Ans: A galvanometer can be converted into an ammeter by connecting a low resistance called shunt parallel to the galvanometer. Since G and ${{R}_{S}}$ are in parallel voltage across then is same ${{I}_{g}}{{R}_{G}}=\left( I-{{I}_{g}} \right){{R}_{s}}$.

${{R}_{s}}=\left( \dfrac{{{I}_{g}}}{I-{{I}_{g}}} \right){{R}_{G}}$

(b) A galvanometer has a resistance 30 and gives a full-scale deflection for a current of 2mA. How much resistance in what way must be connected to convert into?

(i). An ammeter of range 0.3A

Ans: We have:

$I\text{ }=\text{ }0.3A$, $G\text{ }=\text{ }30\Omega$, $Ig\text{ }=\text{ }2mA=2\times {{10}^{-3}}A$

$Shunt\left( S \right)=\dfrac{{{I}_{g}}G}{I-{{I}_{g}}}$

$S=\dfrac{2\times {{10}^{-3}}\times 30}{\left( 0.3-2\times {{10}^{-3}} \right)}=0.2\Omega$

(ii). A voltmeter of range 0.2V.

Ans: We have:

$G\text{ }=\text{ }30\Omega$, $Ig\text{ }=\text{ }2mA=2\times {{10}^{-3}}A$, $V\text{ }=\text{ }0.2V$

$Shu\text{nt Re}sis\tan ce\left( R \right)=\left( \dfrac{V}{{{I}_{g}}}-G \right)$

$\Rightarrow \left( R \right)=\left( \dfrac{0.2}{2\times {{10}^{-3}}}-30 \right)=70\Omega$

6. A monoenergetic ($18keV$) electron beam initially in the horizontal direction is subjected to a horizontal magnetic field of $0.04G$normal to the initial direction. Estimate the up or down deflection of the beam over a distance of $30cm\,\,({{m}_{e}}=9.11\times {{10}^{-31}}kg)$. (Note: Data in this exercise are so chosen that the answer will give you an idea of the effect of earth’s magnetic field on the motion of the electron beam from the electron gun to the screen in a TV set.)

Ans: Provided that,

Energy of an electron beam, $E=18keV=18\times {{10}^{3}}eV$

Charge on an electron, $e=1.6\times {{10}^{-19}}C$

$\therefore E=18\times {{10}^{3}}\times 1.6\times {{10}^{-19}}V=2.88\times {{10}^{-15}}V$

The magnetic field, $B=0.04G$

The mass of an electron, ${{m}_{e}}=9.11\times {{10}^{-31}}kg$

Distance till where the electron beam travels, $d=30cm=0.3m$

We can write the kinetic energy carried by the electron beam as:

$E=\dfrac{1}{2}m{{v}^{2}}$

$\Rightarrow v=\sqrt{\dfrac{2E}{m}}$

$\Rightarrow v=\sqrt{2\times \dfrac{2.88\times {{10}^{-15}}}{9.11\times {{10}^{-31}}}}=0.795\times {{10}^{8}}m/s$

The electron beam deflects and gets in a circular path of radius, $r$ .

The force experienced due to the magnetic field balances the centripetal force of the path.

$BeV=\dfrac{m{{v}^{2}}}{r}$

$\Rightarrow r=\dfrac{mv}{Be}$

$\Rightarrow r=\dfrac{9.11\times {{10}^{-31}}\times 0.795\times {{10}^{8}}}{0.4\times {{10}^{-4}}\times 1.6\times {{10}^{-19}}}=11.3m$

Let the up-down deflection of the beam be $x=r(1-\cos \theta )$.

Where,

$\theta$ is the angle of declination given by,

$\theta ={{\sin }^{-1}}(d/r)=1.521{}^\circ$

And

$x=11.3(1-\cos 1.521{}^\circ )=0.0039m$

Clearly, the up and down deflection of the bean$=3.9mm$.

7. A sample of paramagnetic salt contains $2.0\times {{10}^{24}}$atomic dipoles each of dipole moment $1.5\times {{10}^{-23}}J/T$. The sample is placed under a homogeneous magnetic field of $0.64T$, and cooled to a temperature of $4.2K$ The degree of magnetic saturation achieved is equal to $15%$. What is the total dipole moment of the sample for a magnetic field of $0.98T$and a temperature of $2.8\,K$? (Assume Curie’s law)

Ans: Provided that,

The number of atomic dipoles, $n=2.0\times {{10}^{24}}$

Dipole moment for each atomic dipole, $M=1.5\times {{10}^{-23}}J/T$

The given magnetic field, ${{B}_{1}}=0.64T$

The sample is then cooled to a temperature, ${{T}_{1}}=4.2K$

Total dipole moment of the atomic dipole,

${{M}_{tot}}=n\times M=2\times {{10}^{24}}\times 1.5\times {{10}^{-23}}=30J/T$

Magnetic saturation is achieved at $15%$.

Hence, effective dipole moment, ${{M}_{1}}=\dfrac{15}{100}30=4.5J/T$

Now when the magnetic field is ${{B}_{2}}=0.98T$

Temperature, ${{T}_{2}}=2.8K$

Its total dipole moment $={{M}_{2}}$

According to Curie’s law, the ratio of the two magnetic dipoles at different temperatures is:

$\dfrac{{{M}_{2}}}{{{M}_{1}}}=\dfrac{{{B}_{2}}}{{{B}_{1}}}\dfrac{{{T}_{1}}}{{{T}_{2}}}$

$\Rightarrow {{M}_{2}}=\dfrac{{{B}_{2}}{{T}_{1}}{{M}_{1}}}{{{B}_{1}}{{T}_{2}}}$

$\Rightarrow {{M}_{2}}=10.336J/T$

Clearly, it can be seen that, $10.336J/T$ is the total dipole moment of the sample for a magnetic field of $0.98T$when its temperature is$2.8K$.

8. A telephone cable at a place has four long straight horizontal wires carrying a current of $1.0\,A$ in the same direction east to west. The earth’s magnetic field at the place is$0.39\,G$, and the angle of dip is $35{}^\circ$. The magnetic declination is nearly zero. What are the resultant magnetic fields at points $4.0\,cm$below and above the cable?

Ans: It is provided that,

The number of horizontal wires in the telephone cable, $n=4$

Current in each wire, $I=1.0A$

Earth’s magnetic field at any location, $H=0.39G=0.39\times {{10}^{-4}}T$

The angle of dip at the location, $\delta =35{}^\circ$

and the angle of declination, $\theta \sim0{}^\circ$

For a point that is 4cm below the cable:

Distance, $r=4cm=0.04m$

The horizontal component (parallel to Earth’s Surface) of Earth’s magnetic field is:

${{H}_{H}}=H.\cos \delta -B$

Here, $B$ is the magnetic field at 4 cm due to current I in the four wires and

${{B}_{{}}}=4\dfrac{{{\mu }_{0}}}{2\pi }\dfrac{I}{r}$

Here,  ${{\mu }_{0}}$ is the permeability of free space $=4\pi \times {{10}^{-7}}Tm/A$

$\Rightarrow B=4\dfrac{4\pi \times {{10}^{-7}}}{2\pi }\dfrac{1}{0.01}$

$\Rightarrow B=0.2\times {{10}^{-4}}T=0.2G$

$\therefore {{H}_{H}}=0.39\times \cos 35{}^\circ -0.2\approx 0.12G$

The vertical component (perpendicular to Earth's surface) of earth’s magnetic field is given as:

${{H}_{v}}=H.\sin \delta$

$\Rightarrow {{H}_{v}}=0.39\times \sin 35{}^\circ =0.22G$

The angle between the field with its horizontal component is given as:

$\theta ={{\tan }^{-1}}\dfrac{{{H}_{v}}}{{{H}_{H}}}$

$\Rightarrow \theta ={{\tan }^{-1}}\dfrac{0.22}{0.12}=61.39$

The resultant field at the point is obtained as:

${{H}_{1}}=\sqrt{{{H}_{H}}^{2}+{{H}_{v}}^{2}}$

$\Rightarrow {{H}_{1}}=\sqrt{{{0.22}^{2}}+{{0.12}^{2}}}=0.25G$

For a point that is 4 cm above the cable,

Horizontal component of earth’s magnetic field:

${{H}_{H}}=H.\cos \delta -B$

$\Rightarrow {{H}_{H}}=0.39\,\cos 35{}^\circ +0.2=0.52$

Vertical component of earth’s magnetic field:

${{H}_{v}}=H.\sin \delta$

$\Rightarrow {{H}_{v}}=0.39\,\sin 35{}^\circ =0.22$

The angle $\theta ={{\tan }^{-1}}\dfrac{{{H}_{v}}}{{{H}_{H}}}$

$\Rightarrow \theta ={{\tan }^{-1}}\dfrac{0.22}{0.52}=22.9{}^\circ$

And the resultant field is:

${{H}_{1}}=\sqrt{{{H}_{H}}^{2}+{{H}_{v}}^{2}}$

$\Rightarrow {{H}_{1}}=\sqrt{{{0.22}^{2}}+{{0.52}^{2}}}=0.56G$

Clearly, the resultant magnetic field below the cable is $0.25G$and above the cable is $0.56G$.

(a) Explain qualitatively on the basis of domain picture the irreversibility in the magnetization curve of a ferromagnet.

Ans: The B-H curve i.e., the Hysteresis curve of a ferromagnetic material is as shown in the figure below:

It can be seen from the above-given curve that magnetization-B persists even when the external field-H is removed. This shows the irreversibility of a ferromagnet, i.e., the magnetization will not drop by reducing the magnetization field just the same way it was increased by increasing the magnetization field.

(b) The hysteresis loop of a soft iron piece has a much smaller area than that of a carbon steel piece. If the material is to go through repeated cycles of magnetization, which piece will dissipate greater heat energy?

Ans: The dissipated heat energy is in proportion to the area inside the hysteresis loop. For a carbon steel piece, the hysteresis curve area is large. Thus, it dissipates greater heat energy.

(c) ‘A system displaying a hysteresis loop such as a ferromagnet, is a device for storing memory?’ Explain the meaning of this statement.

Ans: The information of magnetization corresponds to the cycle of magnetization. Also, it can be seen that Hysteresis loops can be used for storing such information.

The value of magnetization is memory or record of hysteresis loop cycles of magnetization.

(d) What kind of ferromagnetic material is used for coating magnetic tapes in a cassette player, or for building ‘memory stores’ in a modern computer?

Ans: Ceramic.

Ceramic is usually used for coating magnetic tapes in memory storage devices like cassette players and also for building memory stores in today's computers.

(e) A certain region of space is to be shielded from magnetic fields. Suggest a method.

Ans: A region of space can be shielded from magnetic fields if it is surrounded by soft iron rings. In such arrangements, the magnetic lines are drawn out of the region.

(a) Why does a paramagnetic sample display greater magnetization (for the same magnetizing field) when cooled?

Ans: The thermal motion of molecules is random, and the randomness increases with increasing temperature. Considering this fact, the alignments of dipoles get disrupted at high temperatures.

On cooling, this disruption is reduced. Hence, a paramagnetic sample displays greater magnetization when its temperature is lowered i.e., it is cooled.

(b) Why is diamagnetism, in contrast, almost independent of temperature?

Ans: In presence of a magnetizing field, the induced dipole moment in a diamagnetic substance is always opposite to the magnetizing field.

Hence, the change in temperature that leads to a change in the internal motion of the atoms does not affect the diamagnetism of a material.

(c) If a toroid uses bismuth for its core, will the field in the core be (slightly) greater or (slightly) less than when the core is empty?

Ans: It is known that Bismuth is a diamagnetic substance. This means, the magnetic field due to the toroid will be the magnetizing field for the bismuth core which will be opposite to the induced magnetic field of Bismuth.

Hence the total field generated by the toroid will be slightly less than the empty-core-toroid.

(d) Is the permeability of a ferromagnetic material independent of the magnetic field? If not, is it more for lower or higher fields?

Ans: The permeability of ferromagnetic materials is dependent of the applied magnetic field. As observed from hysteresis curve, it is greater for a lower field and vice versa.

(d) Magnetic field lines are always nearly normal to the surface of a ferromagnet at every point. (This fact is analogous to the static electric field lines being normal to the surface of a conductor at every point). Why?

Ans: The permeability of ferromagnetic material is greater than 1; not less than 1.  Therefore, magnetic field lines are always nearly normal to the surface of such materials at every point.

(e) Would the maximum possible magnetization of a paramagnetic sample be of the same order of magnitude as the magnetization of a ferromagnet?

Ans: The maximum possible magnetization of a paramagnetic sample can be of the same order of magnitude as the magnetization of a ferromagnet. This requires high magnetizing fields for saturation.

11. A short bar magnet of magnetic moment $5.25\times {{10}^{-2}}J/T$is placed with its axis perpendicular to the earth’s field direction. Magnitude of the earth’s field at the place is given to be $0.42\,G$. Ignore the length of the magnet in comparison to the distances involved.  At what distance from the centre of the magnet, the resultant field is inclined at $45{}^\circ$ with earth’s field on

(a) Its normal bisector and

Ans: Provided that,

The magnetic moment of the bar magnet, $M=5.25\times {{10}^{-2}}J/T$

The magnitude of the Earth’s magnetic field at a place, $G=0.42G=0.42\times {{10}^{-4}}T$

The magnetic field at the distance of R from the centre of the magnet on the normal bisector is given by the relation:

$B=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{M}{{{R}^{3}}}$

Here,

$M$is the above-mentioned magnetic moment

${{\mu }_{0}}$is the permeability of free space

When the resultant field is inclined at $45{}^\circ$ with earth’s field, $B=H$

$\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{M}{{{R}^{3}}}=H=0.42\times {{10}^{-4}}$

$\Rightarrow {{R}^{3}}=\dfrac{4\pi \times {{10}^{-7}}}{4\pi }\dfrac{5.25\times {{10}^{-2}}}{0.42\times {{10}^{-4}}}=12.5\times {{10}^{-5}}$

$\Rightarrow R=5\times {{10}^{-2}}m=5cm$

Clearly, at a distance of 5cm from the centre of the magnet, the resultant field is inclined at $45{}^\circ$ with earth’s field on its normal bisector.

(b) its axis

Ans: Provided that,

The magnetic moment of the bar magnet, $M=5.25\times {{10}^{-2}}J/T$

The magnitude of the Earth’s magnetic field at a place, $G=0.42G=0.42\times {{10}^{-4}}T$

The given magnetic field at $R$ distanced from the centre of the magnet on a point on its axis is given as:

$B'=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2M}{{{R}^{3}}}$

The resultant field is inclined at $45{}^\circ$with earth’s field

$B'=H$

$\Rightarrow \dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2M}{{{(R')}^{3}}}=H$

$\Rightarrow {{(R')}^{3}}=\dfrac{4\pi \times {{10}^{-7}}}{4\pi }\dfrac{2\times 5.25\times {{10}^{-2}}}{0.42\times {{10}^{-4}}}=2.5\times {{10}^{-4}}$

$\Rightarrow R=6.3\times {{10}^{-2}}m=6.3cm$

Clearly, at a distance of 6.3cm from the centre of the magnet, the resultant field is inclined at $45{}^\circ$ with earth’s field on its axis.

12. If the bar magnet in exercise 5.13 is turned around by $180{}^\circ$, where will the new null points be located?

Ans: According to what is given, the magnetic field on the axis of the magnet at a distance ${{d}_{1}}=14cm$, can be written as:

${{B}_{1}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{2M}{{{d}_{1}}^{3}}=H$

here,

$M$is the magnetic moment

${{\mu }_{0}}$is the permeability of free space

$H$ is the horizontal component of the given magnetic field at ${{d}_{1}}$

When the bar magnet is turned through $180{}^\circ$, then the neutral point will lie on the equatorial line.

Also, the magnetic field at a distance ${{d}_{2}}$ on the equatorial line of the magnet can be written as:

${{B}_{2}}=\dfrac{{{\mu }_{0}}}{4\pi }\dfrac{M}{{{d}_{2}}^{3}}=H$

Equating ${{B}_{1}}$and ${{B}_{2}}$we get:

$\dfrac{2}{{{d}_{1}}^{3}}=\dfrac{1}{{{d}_{2}}^{3}}$

$\Rightarrow {{d}_{2}}={{d}_{1}}{{\left( \dfrac{1}{2} \right)}^{1/3}}$

$\Rightarrow 14\times 0.794=11.1cm$

Thus, the new null point will be located $11.1\,cm$ on the normal bisector.

13. A bar magnet of magnetic moment $1.5\,J/T$ lies aligned with the direction of a uniform magnetic field of $0.22\,T$.

(a) What is the amount of work required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?

Ans: Provided that,

Magnetic moment, $M=1.5J/T$

Magnetic field strength, $B=0.22\,T$

(i) Initial angle between the magnetic field and the axis is, ${{\theta }_{1}}=0{}^\circ$

Final angle between the magnetic field and the axis is, ${{\theta }_{2}}=90{}^\circ$

The work that would be required to make the magnetic moment perpendicular to the direction of magnetic field would be:

$W=-MB(\cos {{\theta }_{2}}-\cos {{\theta }_{1}})$

$\Rightarrow W=-1.5\times 0.22(\cos 90{}^\circ -\cos 0{}^\circ )$

$\Rightarrow W=-0.33(0-1)$

$\Rightarrow W=0.33\,J$

(ii) Initial angle between the magnetic field and the axis, ${{\theta }_{1}}=0{}^\circ$

Final angle between the magnetic field and the axis, ${{\theta }_{2}}=180{}^\circ$

The work that would be required to make the magnetic moment opposite (180 degrees) to the direction of magnetic field is given as:

$W=-MB(\cos {{\theta }_{2}}-\cos {{\theta }_{1}})$

$\Rightarrow W=-1.5\times 0.22(\cos 180{}^\circ -\cos 0{}^\circ )$

$\Rightarrow W=-0.33(-1-1)$

$\Rightarrow W=0.66J$

(b) What is the torque on the magnet in cases (i) and (ii)?

Ans: For the first (i) case,

$\theta ={{\theta }_{1}}=90{}^\circ$

Hence the Torque, $\vec{\tau }=\vec{M}\times \vec{B}$

And its magnitude is:$\tau =MB\sin (\theta )$

$\Rightarrow \tau =1.5\times 0.22\sin (90{}^\circ )$

$\Rightarrow \tau =0.33Nm$

Hence the torque involved is $=0.33Nm$

For the second-(ii) case:

$\theta ={{\theta }_{1}}=180{}^\circ$

And its magnitude of the torque is:$\tau =MB\sin (\theta )$

$\Rightarrow \tau =1.5\times 0.22\sin (180{}^\circ )$

$\Rightarrow \tau =0Nm$

Hence the torque is zero.

14. Answer the following questions regarding earth's magnetism:

(a) A vector needs three quantities for its specification. Name the three independent quantities conventionally used to specify the earth's magnetic field.

Ans: The three independent quantities conventionally used for specifying earth's magnetic field are:

(i). Magnetic declination

(iii). Angle of dip

(iii). Horizontal component of earth's magnetic field

(b) The angle of dip at a location in southern India is about ${{18}^{0}}$. Would you expect a greater or smaller dip angle in Britain?

Ans: The angle of dip at a point depends on how far the point is located with respect to the North Pole or the South Pole. The angle of dip would be greater in Britain (it is about $70{}^\circ$) than in southern India because the location of Britain on the globe is closer to the magnetic North Pole.

(c). If you made a map of magnetic field lines at Melbourne in Australia, would the lines seem to go into the ground or come out of the ground?

Ans: It is hypothetically considered that a huge bar magnet is dipped inside earth with its north pole near the geographic South Pole and its south pole near the geographic North Pole. Magnetic field lines emanate from a magnetic north pole and terminate at a magnetic south pole. Hence, in a map depicting earth's magnetic field lines, the field lines at Melbourne, Australia would seem to come out of the ground.

(d) In which direction would a compass free to move in the vertical plane point to, if located right on the geomagnetic north or south pole?

Ans: If a compass is located on the geomagnetic North Pole or South Pole, then the compass will be free to move in the horizontal plane while earth's field is exactly vertical to the magnetic poles. In such a case, the compass can point in any direction.

(e) The earth's field, it is claimed, roughly approximates the field due to a dipole of magnetic moment $8\times {{10}^{22}}J{{T}^{-1}}$ located at its centre. Check the order of magnitude of this number in some way.

Ans: Given that,

Magnetic moment, $M=8\times {{10}^{22}}J{{T}^{-1}}$

Radius of earth, $r=6.4\times {{10}^{6}}m$

Magnetic field strength,

$B=\dfrac{800\times 4\pi \times {{10}^{-7}}\times 1.2\times 3500}{2\pi \times 0.15}=4.48T$

Where,

${{\mu }_{0}}=\text{ }Permeability\text{ }of\text{ }free\text{ }space=4\pi \times {{10}^{-7}}TmA$.

$\Rightarrow \dfrac{4\pi \times {{10}^{-7}}\times 8\times {{10}^{22}}}{4\pi \times {{\left( 6.4\times {{10}^{6}} \right)}^{3}}}=0.3G$

This quantity is of the order of magnitude of the observed field on earth.

(f) Geologists claim that besides the main magnetic N-S poles, there are several local poles on the earth's surface oriented in different directions. How is such a thing possible at all?

Ans: Yes, there are several local poles on earth's surface oriented in different directions. A magnetised mineral deposit is an example of a local N-S pole.

(a) The earth's magnetic field varies from point to point in space. Does it also change with time? If so, on what time scale does it change appreciably?

Ans: Earth's magnetic field changes with time. It takes a few hundred years to change by an appreciable amount. The variation in the earth's magnetic field with time cannot be neglected.

(b) The earth's core is known to contain iron. Yet geologists do not regard this as a source of the earth's magnetism. Why?

Ans: Earth's core contains molten iron. This form of iron is not ferromagnetic. Hence, this is not considered as a source of earth's magnetism.

(c) The charged currents in the outer conducting regions of the earth's core are thought to be responsible for earth's magnetism. What might be the 'battery' (i.e., the source of energy) to sustain these currents?

Ans: The radioactivity in earth's interior is the source of energy that sustains the currents in the outer conducting regions of earth's core. These charged currents are considered to be responsible for earth's magnetism.

(d) The earth may have even reversed the direction of its field several times during its history of 4 to 5 billion years. How can geologists know about the earth's field in such distant past?

Ans: Earth reversed the direction of its field several times during its history of $4$ to $5$ billion years. These magnetic fields got weakly recorded in rocks during their solidification. One can get clues about the geomagnetic history from the analysis of this rock magnetism.

(e) The earth's field departs from its dipole shape substantially at large distances (greater than about 30,000 km). What agencies may be responsible for this distortion?

Ans: Earth's field departs from its dipole shape substantially at large distances (greater than about $30,000$ km) because of the presence of the ionosphere. In this region, earth's field gets modified because of the field of single ions. While in motion, these ions produce the magnetic field associated with them.

(f) Interstellar space has an extremely weak magnetic field of the order of 10-12 T. Can such a weak field be of any significant consequence? Explain. (Note: Exercise 5.2 is meant mainly to arouse your curiosity. Answers to some questions above are tentative or unknown. Brief answers wherever possible are given at the end. For details, you should consult a good text on geomagnetism.)

Ans: An extremely weak magnetic field can bend charged particles moving in a circle. This may not be noticeable for a large radius path. With reference to the gigantic interstellar space, the deflection can affect the passage of charged particles.

## Important Questions For Class 12 Physics Chapter 5

### Magnetism And Matter Class 12 Important Questions

Chapter 5 of class 12 physics deals with the concepts related to Magnetism and Matter. The magnetism and matter class 12 important questions PDF will help the students in boosting their exam preparation. Chapter 5 of class 12 physics will cover good marks in the CBSE board exams. Students are advised to cover all the topics and subtopics with proper study materials like important questions for class 12 physics chapter 5 PDF available on the Vedantu website. All the study materials are available in Pdf format for easy access and students can download these anytime.

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• Chapter 5 begins with the introduction to the idea of magnetism and matter.

• We come across important concepts like determining the magnetic field and force. The derivation of the Lorentz force will be covered.

• This chapter will learn a few important derivations like Biot Savart’s law, Gauss law, etc.

• Construction and working of moving coil galvanometer, Cyclotrons, etc.. are studied in detail.

• We will see about the magnetic moment and determination of the magnetic moment. Students will be learning important concepts like magnetization and magnetic intensity.

• Magnetic properties of materials and their classification. We learn about what are diamagnetic, paramagnetic, and ferromagnetic properties and the difference between them.

The magnetism and matter chapter gives a strong basic understanding which will be useful while studying advanced concepts of the modern physics chapter. These are the few points that have to be remembered while solving magnetism and matter class 12 important questions PDF. This topic gives a good understanding of the magnetic properties of the different materials.

### Conclusion:

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## FAQs on Important Questions for CBSE Class 12 Physics Chapter 5 - Magnetism and Matter 2024-25

1. Is it possible to download the important questions of Chapter 5 of Class 12 Physics?

Below are steps that will help to download the important questions of Chapter 5 of Class 12 Physics:

• Tap on the provided link-Important questions of class 12 physics.

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2. How to score well in Chapter 5 of Class 12 Physics?

The following strategy will be helpful in scoring good marks in Chapter 5 of Class 12 Physics:

• Read Chapter 5 thoroughly from the NCERT Biology book as the content is based on the CBSE curriculum.

• Solve each question of the chapter from the NCERT book.

• Practice previous years question papers and sample papers to get an idea about the type of questions asked in the exam.

• Make use of guidebooks to understand the concepts of the chapter more easily.

3. What are the various terminologies related to bar magnets?

• Magnetic Pole –The endpoints of the bar magnet where the magnetism is maximum is known as magnetic poles. These are of two types, the North pole and the South Pole.

• Magnetic Axis –It can be defined as the imaginary line which joins the magnetic poles of the bar magnet.

• Effective Length –The effective length is the length between the poles of the magnet. It is 5/6th part of the actual length of the bar magnet.

• Pole Strength –It is termed as the ability of the magnetic poles to attract magnetic materials.

4. Define magnetic dipole.

A magnetic dipole is defined as the pair of the equal magnitude of pole strength but opposite nature of poles separated by a small distance “2l”.

Mathematically, the magnetic moment is

M = m * 2l

• “Ampere metre square” is the SI unit of the magnetic moment.

• This is a type of vector quantity.

• Direction of the magnetic moment is from North to South.

• If the direction of current is clockwise then the direction of the magnetic moment is inwards.

• If the direction of current is anti-clockwise then the direction of magnetic moment is outwards.

5. Explain the following:

• Magnetic Flux Density

• Intensity Of Magnetisation

• Magnetic Permeability

• Magnetic Flux Density – It is defined as the number of magnetic field lines passing through the unit area of the surface.

The SI unit of magnetic flux is “Tesla” and is denoted by “B”.

• Intensity of Magnetisation –The amount or extent or degree to which a substance is magnetised when placed in a magnetic field is known as the intensity of magnetisation.

It is denoted by “I” and the SI unit is “ampere per metre”. The magnetisation intensity is a vector quantity.

• Magnetic Permeability – The extent to which magnetic field lines can enter a substance is called magnetic permeability and the SI unit is “Tesla-metre per ampere”.