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NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges And Magnetism

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NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges and Magnetism

The NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges and Magnetism provided by Vedantu is a comprehensive and detailed explanation of the concepts covered in the PDF which is FREE to Download! This topic deals with the interaction between moving charges and magnetic fields. In the given PDF the chapter begins by introducing the concept of the magnetic field and its properties. It then discusses the Lorentz force, which is the force exerted on a moving charge by a magnetic field. The chapter also covers the Biot-Savart law, which gives the magnetic field produced by a current element. Ampere's circuital law is also discussed, which gives the relationship between the magnetic field and the current enclosed by a loop. The solutions are also accompanied by solved examples and practice questions, which help students to understand the concepts better.


Class:

NCERT Solutions for Class 12

Subject:

Class 12 Physics

Chapter Name:

Chapter 4 - Moving Charges and Magnetism

Content-Type:

Text, Videos, Images and PDF Format

Academic Year:

2023-24

Medium:

English and Hindi

Available Materials:

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  • Important Questions

  • Revision Notes

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Moving Charges and Magnetism Chapter at a Glance - Class 12 NCERT Solutions

  • The total force on a charge q moving with velocity v in the presence of magnetic and electric fields B and E, respectively is called the Lorentz force. It is given by the expression: F = q (v × B + E). The magnetic force q (v × B) is normal to v and work done by it is zero.

  • A straight conductor of length l and carrying a steady current I experiences a force F in a uniform external magnetic field B, $F=I\left ( l\times B \right )$ 

Where |l| = l and the direction of l is given by the direction of the current.

  • In a uniform magnetic field B, a charge q executes a circular orbit in a plane normal to B. Its frequency of uniform circular motion is called the cyclotron frequency and is given by: 

$v_e=\frac{qB}{2\pi m}$

  • This frequency is independent of the particle’s speed and radius. This fact is exploited in a machine, the cyclotron, which is used to accelerate charged particles. 

Magnetic force does not work when the charged particle is displaced while electric force does work in displacing the charged particle.


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Cyclotrons cannot accelerate electrons because they have very small mass.

  • The Biot-Savart law asserts that the magnetic field dB due to an element dl carrying a steady current I at a point P at a distance r from the current element is:

$dB=\frac{\mu_0}{4\pi}\mathbb{I}\frac{\vec{dl}\times \vec{r}}{r^3}$

To obtain the total field at P, we must integrate this vector expression over the entire length of the conductor.


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Magnetic Field at Centre O in different conditions  

Condition

Figure

Magnetic Field

Are subtends angle at $\theta$ the centre 

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Are subtends angle $(2^{\pi }-^{\theta })$

at the centre

Shape

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Semi – circular are 

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Three quarter semi  circular current carrying are 

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Circular current carrying are

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Concentric co – planer circular loops carries current in the same direction 

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Concentric co-planer circular loops carries current in the opposite direction 

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Concentric loops but their planes are perpendicular to each other 

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Concentric loops but their planes are at an angle $\theta$ with each other 

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Distribution of current across the diameter 

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B = 0 

Distribution of current between any two points on the circumference 

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B = 0 


  • Ampere’s Circuital Law:  Let an open surface S be bounded by a loop C. Then the Ampere’s law states that $\underset{C}{\oint }N/d=\mu_0I$ where Ι refers to the current passing through S. The sign of I is determined from the right-hand rule. If B is directed along the tangent to every point on the perimeter L of a closed curve and is constant in magnitude along perimeter then:       $ BL=\mu_0I_e$ , where Ie is the net current enclosed by the closed circuit

  • The magnitude of the field B inside a long solenoid carrying a current I is: B = $\mu$0nl

where n is the number of turns per unit length. For a toroid one obtains, $B=\frac{\mu_0NI}{2\pi R}$ , where N is the total number of turns and r is the average radius.


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  • If a current carrying circular loop (n = 1) is turned into a coil having n identical turns then magnetic field at the centre of the coil becomes n2 times the previous field i.e. B(n turn) = n2 B(single turn).

  • Parallel currents attract and anti-parallel currents repel.

  • A planar loop carrying a current I, having N closely wound turns, and an area A possesses a magnetic moment M where, M = N I A and the direction of M is given by the right-hand thumb rule.

When this loop is placed in a uniform magnetic field B, the force F on it is:  F = 0

And the torque on it is: $\tau =M\times B$ 

In a moving coil galvanometer, this torque is balanced by a counter- torque due to a spring, yielding: $k\varphi =NIAB$

  • An electron moving around the central nucleus has a magnetic moment M given by: $M=\frac{e}{2M}L$ 

where L is the magnitude of the angular momentum of the circulating electron about the central nucleus and m is the mass. The smallest value of M is called the Bohr magneton MB and it is MB = 9.27×10–24 J/T.

Access NCERT Solutions for Class 12 Physics Chapter 4 – Moving Charges and Magnetism

1. A Circular Coil of Wire Consisting of $100$ Turns, Each of Radius $8.0cm$ Carries a Current of $0.40A$. What is the Magnitude of the Magnetic Field B at the Centre of the Coil?

Ans: We are given:

Number of turns on the circular coil, $n=100$

Radius of each turn, $r=8.0cm=0.08m$

Current flowing in the coil is given to be, $I=0.4A$

We know the expression for magnetic field at the centre of the coil as, 

$\left| B \right|=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi nI}{r}$

Where, ${{\mu }_{0}}=4\pi \times {{10}^{-4}}Tm{{A}^{-1}}$is the permeability of free space. 

On substituting the given values we get, 

$\left| B \right|=\frac{4\pi \times {{10}^{-7}}\times 2\pi \times 100\times 0.4}{4\pi \times 0.08}$

$\Rightarrow \left| B \right|=3.14\times {{10}^{-4}}T$

Clearly, the magnitude of the magnetic field is found to be $3.14\times {{10}^{-4}}T$. 


2. A Long Straight Wire Carries a Current of $35A$. What Is the Magnitude of Field B at a Point 20cm from the Wire?

Ans: We are given the following:

Current in the wire, $I=35A$

Distance of the given point from the wire, $r=20cm=0.2m$

We know the expression for magnetic field as,

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

Where, ${{\mu }_{0}}=4\pi \times {{10}^{-4}}Tm{{A}^{-1}}$is the permeability of free space. 

On substituting the given values, we get, 

$B=\frac{4\pi \times {{10}^{-7}}\times 2\times 35}{4\pi \times 0.2}$

$\Rightarrow B=3.5\times {{10}^{-5}}T$

Thus, we found the magnitude of the magnetic field at the given point to be $3.5\times {{10}^{-5}}T$.


3. A Long Straight Wire in the Horizontal Plane Carries a Current of $50A$ in North to South Direction. Give the Magnitude and Direction of B at a Point $2.5m$ East of the Wire. 

Ans: We are given the following:

The current in the wire, $I=50A$

The distance of the given point from the wire, $r=2.5m$


A long straight wire carrying current in N-S direction


We have the expression for magnetic field as, 

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

Where, ${{\mu }_{0}}=4\pi \times {{10}^{-4}}Tm{{A}^{-1}}$is the permeability of free space. 

Substituting the given values, we get,

$B=\frac{4\pi \times {{10}^{-7}}\times 2\times 50}{4\pi \times 2.5}$

$\Rightarrow B=4\times {{10}^{-6}}T$

Now from Maxwell’s right hand thumb rule, we have the direction of the magnetic field at the given point B to be vertically upward. 


4. A Horizontal Overhead Power Line Carries a Current of $90A$ in East to West Direction. What is the Magnitude and Direction of the Magnetic Field Due to the Current $1.5m$ Below the Line? 

Ans: We are given the following:

Current in the power line, $I=90A$

Distance of the mentioned point below the power line, $r=1.5m$

Now, we have the expression for magnetic field as, 

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

Where, ${{\mu }_{0}}=4\pi \times {{10}^{-4}}Tm{{A}^{-1}}$is the permeability of free space. 

On substituting the given values, we get, 

$B=\frac{4\pi \times {{10}^{-7}}\times 2\times 90}{4\pi \times 1.5}$

$\Rightarrow B=1.2\times {{10}^{-5}}T$

We found the magnitude of the magnetic field to be $1.2\times {{10}^{-5}}T$and it will be directed towards south as per Maxwell’s right hand thumb rule. 


5. What is the Magnitude of Magnetic Force Per Unit Length on a Wire Carrying a Current of 8 and Making an Angle of $30{}^\circ $ with the Direction of a Uniform Magnetic Field of $0.15T$? 

Ans: Given that,

Current in the wire, $I=8A$

Magnitude of the uniform magnetic field, $B=0.15T$

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

We have the expression for magnetic force per unit length on the wire as, 

$F=BI\sin \theta $

Substituting the given values, we get,

$F=0.15\times 8\times 1\times \sin 30{}^\circ $

$\Rightarrow F=0.6N{{m}^{-1}}$

Thus, the magnetic force per unit length on the wire is found to be $0.6N{{m}^{-1}}$


6. A $3.0cm$ Wire Carrying a Current of $10A$ is Placed Inside a Solenoid Perpendicular to Its Axis. The Magnetic Field Inside the Solenoid Is Given to Be $0.27T$. What is the Magnetic Force on the Wire?

Ans: We are given the following, 

Length of the wire, $l=3cm=0.03m$

Current flowing in the wire, $I=10A$

Magnetic field, $B=0.27T$

Angle between the current and magnetic field, $\theta =90{}^\circ $

(Since the magnetic field produced by a solenoid is along its axis and current carrying wire is kept perpendicular to the axis) 

The magnetic force exerted on the wire is given as, 

$F=BIl\sin \theta $

Substituting the given values, 

$F=0.27\times 10\times 0.03\sin 90{}^\circ $

$\Rightarrow F=8.1\times {{10}^{-2}}N$

Clearly, the magnetic force on the wire is found to be $8.1\times {{10}^{-2}}N$. The direction of the force can be obtained from Fleming’s left-hand rule. 


7. Two Long and Parallel Straight Wires A and B Carrying Currents of $8.0A$and $5.0A$ in the Same Direction are Separated by a Distance of $4.0cm$. Estimate the Force on a $10cm$ Section of Wire A. 

Ans: We are given:

Current flowing in wire A, ${{I}_{A}}=8.0A$ 

Current flowing in wire B, ${{I}_{B}}=5.0A$

Distance between the two wires, $r=4.0cm=0.04m$

Length of a section of wire A, $l=10cm=0.1m$

Force exerted on length $l$ due to the magnetic field is given as,

$B=\frac{2{{\mu }_{0}}{{I}_{A}}{{I}_{B}}l}{4\pi r}$ 

Where, ${{\mu }_{0}}=4\pi \times {{10}^{-4}}Tm{{A}^{-1}}$is the permeability of free space. 

On substituting the given values, we get, 

$B=\frac{4\pi \times {{10}^{-7}}\times 2\times 8\times 5\times 0.1}{4\pi \times 0.04}$

$\Rightarrow B=2\times {{10}^{-5}}N$

The magnitude of force is $2\times {{10}^{-5}}N$. This is an attractive force that is normal to A towards B because the direction of the currents in the wires is the same.


8. A Closely Wound Solenoid $80cm$ Long has $5$ Layers of Windings of $400$ Turns Each. The Diameter of the Solenoid is $1.8cm$. If the Current Carried is $8.0A$, Estimate the Magnitude of B Inside the Solenoid Near its Centre. 

Ans: We are given the following:

Length of the solenoid, $l=80cm=0.8m$

Since there are five layers of windings of 400 turns each on the solenoid. 

Total number of turns on the solenoid would be, $N=5\times 400=2000$

Diameter of the solenoid, $D=1.8cm=0.018m$

Current carried by the solenoid, $I=8.0A$

We have the magnitude of the magnetic field inside the solenoid near its centre given by the relation, 

$B=\frac{{{\mu }_{0}}NI}{l}$

Where, ${{\mu }_{0}}=4\pi \times {{10}^{-4}}Tm{{A}^{-1}}$is the permeability of free space. 

On substituting the given values we get, 

$B=\frac{4\pi \times {{10}^{-7}}\times 2000\times 8}{0.8}$

$\Rightarrow B=2.512\times {{10}^{-2}}T$

Clearly, the magnitude of the magnetic field inside the solenoid near its centre is found to be $2.512\times {{10}^{-2}}T$.


9. A Square Coil of Side $10cm$ Consists of 20 Turns and Carries a Current of $12A$. The Coil Is Suspended Vertically and the Normal to the Plane of the Coil Makes an Angle of $30{}^\circ $ with the Direction of a Uniform Horizontal Magnetic Field of Magnitude $0.80T$. What is the Magnitude of Torque Experienced by the Coil?

Ans: We are given the following:

Length of a side of the square coil, $l=10cm=0.1m$

Area of the square, $A={{l}^{2}}={{\left( 0.1 \right)}^{2}}=0.01{{m}^{2}}$

Current flowing in the coil, $I=12A$

Number of turns on the coil, $n=20$

Angle made by the plane of the coil with magnetic field, $\theta =30{}^\circ $

Strength of magnetic field, $B=0.80T$

Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation,

$\tau =nIAB\sin \theta $

Substituting the given values, we get,

$\tau =20\times 0.8\times 12\times 0.01\times \sin 30{}^\circ $

$\Rightarrow \tau =0.96Nm$

Thus, the magnitude of the torque experienced by the coil is 0.96 N m.


10. Two Moving Coil Meters, ${{M}_{1}}$ and ${{M}_{2}}$ Have the Following Particulars:

${{R}_{1}}=10\Omega $  , ${{N}_{1}}=30$, ${{A}_{1}}=3.6\times {{10}^{-3}}{{m}^{2}}$ , ${{B}_{1}}=0.25T$,${{R}_{2}}=14\Omega $ ,${{N}_{2}}=42$${{A}_{2}}=1.8\times {{10}^{-3}}{{m}^{2}}$ , ${{B}_{2}}=0.50T$

(The spring constants are identical for the meters).

Determine the Ratio of:

a) Current Sensitivity of ${{M}_{2}}\text{ and }{{\text{M}}_{1}}$

Ans: We are given:

For moving coil meter ${{M}_{1}}$,

Resistance, ${{R}_{1}}=10\Omega $

Number of turns, ${{N}_{1}}=30$

Area of cross-section, ${{A}_{1}}=3.6\times {{10}^{-3}}{{m}^{2}}$

Magnetic field strength, ${{B}_{1}}=0.25T$

Spring constant, ${{K}_{1}}=K$

For moving coil meter ${{M}_{2}}$:

Resistance, ${{R}_{2}}=14\Omega $

Number of turns, ${{N}_{2}}=42$

Area of cross-section, ${{A}_{2}}=1.8\times {{10}^{-3}}{{m}^{2}}$

Magnetic field strength, ${{B}_{2}}=0.50T$

Spring constant, ${{K}_{2}}=K$

Current sensitivity of ${{M}_{1}}$ is given as:

${{I}_{S1}}=\frac{{{N}_{1}}{{B}_{1}}{{A}_{1}}}{{{K}_{1}}}$

And, current sensitivity of ${{M}_{2}}$ is given as:

${{I}_{S2}}=\frac{{{N}_{2}}{{B}_{2}}{{A}_{2}}}{{{K}_{2}}}$

On taking the ratio, we get, 

$\Rightarrow \frac{{{I}_{S2}}}{{{I}_{S1}}}=\frac{\frac{{{N}_{2}}{{B}_{2}}{{A}_{2}}}{{{K}_{2}}}}{\frac{{{N}_{1}}{{B}_{1}}{{A}_{1}}}{{{K}_{1}}}}$

Substituting the values we get,

$\Rightarrow \frac{{{I}_{S2}}}{{{I}_{S1}}}=\frac{42\times 0.5\times 1.8\times {{10}^{-3}}\times 10\times K}{14\times 30\times 0.25\times 3.6\times {{10}^{-3}}\times K}$

$\Rightarrow \frac{{{I}_{S2}}}{{{I}_{S1}}}=1.4$

Therefore, the ratio of current sensitivity of ${{M}_{2}}\text{ and }{{\text{M}}_{1}}$ is 1.4.


b) Voltage Sensitivity of ${{M}_{2}}\text{ and }{{\text{M}}_{1}}$

Ans: Voltage sensitivity for ${{M}_{2}}$is given is:

${{V}_{S2}}=\frac{{{N}_{2}}{{B}_{2}}{{A}_{2}}}{{{K}_{2}}{{R}_{2}}}$

And, voltage sensitivity for ${{M}_{1}}$is given as:

${{V}_{S1}}=\frac{{{N}_{1}}{{B}_{1}}{{A}_{1}}}{{{K}_{1}}{{R}_{1}}}$

On taking the ratio we get, 

$\Rightarrow \frac{{{V}_{S2}}}{{{V}_{S1}}}=\frac{{{N}_{2}}{{B}_{2}}{{A}_{2}}{{K}_{1}}{{R}_{1}}}{{{K}_{2}}{{R}_{2}}{{N}_{1}}{{B}_{1}}{{A}_{1}}}$

Substituting the given values, we get, 

$\Rightarrow \frac{{{V}_{S2}}}{{{V}_{S1}}}=\frac{42\times 0.5\times 1.8\times {{10}^{-3}}\times 10\times K}{K\times 14\times 30\times 0.25\times 3.6\times {{10}^{-3}}}=1$

Thus, the ratio of voltage sensitivity of ${{M}_{2}}\text{ and }{{\text{M}}_{1}}$is 1. 


11. In a Chamber, a Uniform Magnetic Field of $6.5G\left( 1G={{10}^{-4}}T \right)$is Maintained. An Electron Is Shot Into the Field With a Speed Of $4.8\times {{10}^{6}}m{{s}^{-1}}$ Normal to the Field. Explain Why the Path of the Electron is a Circle. Determine the Radius of the Circular Orbit.$\left( e=1.6\times {{10}^{-19}}C,{{m}_{e}}=9.1\times {{10}^{-31}}kg \right)$

Ans: Magnetic field strength, $B=6.5G=6.5\times {{10}^{-4}}T$

Speed of the electron, $V=4.8\times {{10}^{6}}m/s$

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

Mass of the electron, ${{m}_{e}}=9.1\times {{10}^{-31}}kg$

Angle between the shot electron and magnetic field, $\theta =90{}^\circ $

Magnetic force exerted on the electron in the magnetic field could be given as:

$F=evB\sin \theta $

This force provides centripetal force to the moving electron and hence, the electron starts moving in a circular path of radius r.

Hence, centripetal force exerted on the electron would be,

${{F}_{C}}=\frac{m{{v}^{2}}}{r}$

However, we know that in equilibrium, the centripetal force exerted on the electron is equal to the magnetic force i.e.,

${{F}_{C}}=F$

$\Rightarrow \frac{m{{v}^{2}}}{r}=evB\sin \theta $

$\Rightarrow r=\frac{mv}{Be\sin \theta }$

Substituting the given values we get, 

$\Rightarrow r=\frac{9.1\times {{10}^{-31}}\times 4.8\times {{10}^{6}}}{6.5\times {{10}^{-4}}\times 1.6\times {{10}^{-19}}\times \sin 90{}^\circ }$

$\Rightarrow r=4.2cm$

Clearly, we found the radius of the circular orbit to be 4.2cm.


12. In Exercise 4.11 Obtain the Frequency of Revolution of the Electron in Its Circular Orbit. Does the Answer Depend on the Speed of the Electron? Explain.

Ans: We are given the following:

Magnetic field strength, $B=6.5\times {{10}^{-4}}T$

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

Mass of the electron, ${{m}_{e}}=9.1\times {{10}^{-31}}kg$

Velocity of the electron, $v=4.8\times {{10}^{6}}m/s$

Radius of the orbit, $r=4.2cm=0.042m$

Frequency of revolution of the electron $\nu $

Angular frequency of the electron $\omega =2\pi \theta $

Velocity of the electron is related to the angular frequency as:

$v=r\omega $

In the circular orbit, the magnetic force on the electron provides the centripetal force. Hence,

$evB=\frac{m{{v}^{2}}}{r}$

$\Rightarrow eB=\frac{m}{r}\left( r\omega  \right)=\frac{m}{r}\left( r2\pi \nu  \right)$

$\Rightarrow \nu =\frac{6.5\times {{10}^{-4}}\times 1.6\times {{10}^{-19}}}{2\times 3.14\times 9.1\times {{10}^{-31}}}$$\therefore \nu =18.2\times {{10}^{6}}Hz\approx 18MHz$

Thus, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.


13. 

a) A Circular Coil of 30 Turns and Radius $8.0cm$ Carrying a Current of $6.0A$ is Suspended Vertically in a Uniform Horizontal Magnetic Field of Magnitude $1.0T$ The Field Lines Make an Angle of $60{}^\circ $ with the Normal of the Coil. Calculate the Magnitude of the Counter Torque That Must Be Applied to Prevent the Coil from Turning.

Ans: Number of turns on the circular coil, $n=30$

Radius of the coil, $r=8.0cm=0.08m$

Area of the coil, $A=\pi {{r}^{2}}=\pi {{\left( 0.08 \right)}^{2}}=0.0201{{m}^{2}}$

Current flowing in the coil is given to be,  $I=6.0A$

Magnetic field strength, $B=1T$

Angle between the field lines and normal with the coil surface, $\theta =60{}^\circ $

The coil will turn when it experiences a torque in the magnetic field. The counter torque applied to prevent the coil from turning is given by the relation, $\tau =nIAB\sin \theta $

$\Rightarrow \tau =30\times 6\times 1\times 0.0201\times \sin 60{}^\circ $

$\Rightarrow \tau =3.133Nm$

b) Would Your Answer Change, If the Circular Coil in (a) Were Replaced by a Planar Coil of Some Irregular Shape that Encloses the Same Area? (all Other Particulars Are Also Unaltered).

Ans: From the part(a) we could infer that the magnitude of the applied torque is not dependent on the shape of the coil. 

On the other hand, it is dependent on the area of the coil. 

Thus, we could say that the answer would not change if the circular coil in the above case is replaced by a planar coil of some irregular shape that encloses the same area.


14. Two Concentric Circular Coils X and Y Radii 16 cm and 10 cm, Respectively, Lie in the Same Vertical Plane Containing the North to South Direction. Coil X has 20 Turns and Carries a Current of 16 A; Coil Y has 25 Turns and Carries a Current Of 18 A. The Sense of the Current in X is Anticlockwise, and Clockwise in Y, for an Observer Looking at the Coils Facing West. Give the Magnitude and Direction of the Net Magnetic Field Due to the Coils at Their Centre.

Ans: We are given,

Radius of coil X, ${{r}_{1}}=16cm=0.16m$

Radius of coil Y, ${{r}_{2}}=10cm=0.1m$

Number of turns of on coil X,${{n}_{1}}=20$

Number of turns of on coil Y,${{n}_{2}}=25$

Current in coil X, ${{I}_{1}}=16A$

Current in coil Y,${{I}_{2}}=18A$

Magnetic field due to coil X at their centre is given by the relation,

${{B}_{1}}=\frac{{{\mu }_{0}}{{n}_{1}}{{I}_{1}}}{2{{r}_{1}}}$

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

${{B}_{1}}=\frac{4\pi \times {{10}^{-7}}\times 20\times 16}{2\times 0.16}$

$\Rightarrow {{B}_{1}}=4\pi \times {{10}^{-4}}T$(towards East)

Magnetic field due to coil Y at their centre is given by the relation,

${{B}_{2}}=\frac{{{\mu }_{0}}{{n}_{2}}{{I}_{2}}}{2{{r}_{2}}}$

$\Rightarrow {{B}_{2}}=\frac{4\pi \times {{10}^{-7}}\times 25\times 18}{2\times 0.10}$

$\Rightarrow {{B}_{2}}=9\pi \times {{10}^{-4}}T$(towards West)

Clearly, net magnetic field could be obtained as:

$B={{B}_{2}}-{{B}_{1}}=9\pi \times {{10}^{-4}}-4\pi \times {{10}^{-4}}$

$\Rightarrow B=1.57\times {{10}^{-3}}T$(towards West)


15. A magnetic field of $100G$$\left( where,\text{ }1G={{10}^{-4}}T \right)$ is required which is uniform in a region of linear dimension about $10cm$ and area of cross-section about${{10}^{-3}}{{m}^{2}}$. The maximum current carrying capacity of a given coil of wire is $15A$ and the number of turns per unit length that can be wound a core is at most $1000\text{ turns per m}$. Suggest some appropriate design particulars to a solenoid for the required purpose. Assume the core is not ferromagnetic.

Ans: We are given,

Magnetic field strength,$B=100G=100\times {{10}^{-4}}T$

Number of turns per unit length,$n=1000turns\text{ per m}$

Current flowing in the coil,$I=15A$

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

Magnetic field is given the relation,

$B={{\mu }_{0}}nI$

$\Rightarrow nI=\frac{B}{{{\mu }_{0}}}=\frac{100\times {{10}^{-4}}}{4\pi \times {{10}^{-7}}}$

$\Rightarrow nI\approx 8000A/m$

If the length of the coil is taken as 50 cm, radius 4 cm, number of turns 400, and current 10 A, then these values are not unique for the given purpose. There is always a possibility of some adjustments with limits.


16. For a Circular Coil of Radius R and N Turns Carrying Current I, the Magnitude of the Magnetic Field at a Point on Its Axis at a Distance X from Its Centre Is Given By,

a) Show that this reduces to the familiar result for the field at the centre of the coil.

Ans: We are given,

Radius of circular coil = R

Number of turns on the coil = N

Current in the coil = I

Magnetic field at a point on its axis at distance x is given by the relation,

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

If the magnetic field at the centre of the coil is considered, then $x=0$

$B=\frac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\frac{3}{2}}}}$

This is the familiar result for the magnetic field at the centre of the coil.

b) Consider Two Parallel Co-axial Circular Coils of Equal Radius R, and the Number of Turns N, Carrying Equal Currents in the Same Direction, and Separated by a Distance R. Show That the Field on the Axis Around the Mid-point Between the Coils Is Uniform Over a Distance That Is Small as Compared to R, and Is Given By, Approximately, (such an Arrangement to Produce a Nearly Uniform Magnetic Field Over a Small Region Is Known as Helmholtz Coils.)

Ans: Given that,

Radii of two parallel co-axial circular coils = R

Number of turns on each coil = N

Current in both coils = I

Distance between both the coils = R

Let us consider point Q at distance d from the centre.

Then, one coil is at a distance of $\frac{R}{2}+d$from point Q.

Magnetic field at point Q could be given as:

$B=\frac{{{\mu }_{0}}I{{R}^{2}}N}{2{{\left( {{x}^{2}}+{{R}^{2}} \right)}^{\frac{3}{2}}}}$

Also, the other coil is at a distance of  $\frac{R}{2}+d$from point Q.

Magnetic field due to this coil is given as:

${{B}_{2}}=\frac{{{\mu }_{0}}NI{{R}^{2}}}{2{{\left[ {{\left( \frac{R}{2}-d \right)}^{2}}+{{R}^{2}} \right]}^{\frac{3}{2}}}}$

Now we have the total magnetic field as, 

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

$\Rightarrow B=\frac{{{\mu }_{0}}I{{R}^{2}}}{2}\left[ \left\{ {{\left\{ \left. {{\left( \frac{R}{2}-d \right)}^{2}}+{{R}^{2}} \right\} \right.}^{\frac{-3}{2}}}+{{\left. \left\{ {{\left( \frac{R}{2}+d \right)}^{2}}+{{R}^{2}} \right. \right\}}^{\frac{-3}{2}}}\times N \right. \right]$

$\Rightarrow B=\frac{{{\mu }_{0}}I{{R}^{2}}}{2}\left[ \left\{ {{\left\{ \left. \frac{5{{R}^{2}}}{4}+{{d}^{2}}-Rd \right\} \right.}^{\frac{-3}{2}}}+{{\left. \left\{ \frac{5{{R}^{2}}}{4}+{{d}^{2}}+Rd \right. \right\}}^{\frac{-3}{2}}}\times N \right. \right]$

$\Rightarrow B=\frac{{{\mu }_{0}}I{{R}^{2}}}{2}\left[ \left\{ {{\left\{ \left. 1+\frac{4{{d}^{2}}}{5{{R}^{2}}}-\frac{4d}{5R} \right\} \right.}^{\frac{-3}{2}}}+{{\left\{ 1+\frac{4{{d}^{2}}}{5{{R}^{2}}}+\frac{4d}{5R} \right\}}^{\frac{-3}{2}}}\times N \right. \right]$

Now for $d\ll R$, we could neglect the factor $\frac{{{d}^{2}}}{{{R}^{2}}}$, we get, 

$B\approx \frac{{{\mu }_{0}}I{{R}^{2}}}{2}\times {{\left( \frac{5{{R}^{2}}}{4} \right)}^{\frac{-3}{2}}}\left[ {{\left( 1-\frac{4d}{5R} \right)}^{\frac{-3}{2}}}+{{\left( 1+\frac{4d}{5R} \right)}^{\frac{-3}{2}}} \right]\times N$

$\Rightarrow B\approx \frac{{{\mu }_{0}}I{{R}^{2}}}{2}\times {{\left( \frac{5{{R}^{2}}}{4} \right)}^{\frac{-3}{2}}}\left[ 1-\frac{6d}{5R}+1+\frac{6d}{5R} \right]$

$\Rightarrow B\approx {{\left( \frac{4}{5} \right)}^{\frac{3}{2}}}\frac{{{\mu }_{0}}IN}{R}=0.72\left( \frac{{{\mu }_{0}}IN}{R} \right)$

Clearly, we proved that the field along the axis around the mid-point between the coils is uniform. 


17. A Toroid Has a Core (non-Ferromagnetic) of Inner Radius 25 Cm and Outer Radius 26 Cm, Around Which 3500 Turns of a Wire Are Wound. If the Current in the Wire is 11 A, What is the Magnetic Field 

a) Outside the Toroid 

Ans: We are given,

Inner radius of the toroid, ${{r}_{1}}=25cm=0.25m$

Outer radius of the toroid, ${{r}_{2}}=26cm=0.26m$

Number of turns on the coil, $N=3500$

Current in the coil, $I=11A$

Magnetic field outside a toroid is zero. It is non-zero only inside the core of a toroid.


b) Inside the Core of the Toroid.

Ans: Magnetic field inside the core of a toroid is given by the relation, $B=\frac{{{\mu }_{0}}NI}{l}$

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

$l$ is the length of toroid, given by

$l=2\pi \left( \frac{{{r}_{1}}+{{r}_{2}}}{2} \right)=\pi \left( 0.25+0.26 \right)=0.51\pi $

$\Rightarrow B=\frac{4\pi \times {{10}^{-7}}\times 3500\times 11}{0.51\pi }\approx 3.0\times {{10}^{-2}}T$

Thus, magnetic field inside the core of the toroid is $3.0\times {{10}^{-2}}T$.


c) In the Empty Space Surrounded by the Toroid.

Ans: The empty space that is surrounding the toroid has a magnetic field equal to zero. 


18. Answer the Following Questions:

a) A Magnetic Field That Varies in Magnitude from Point to Point but Has a Constant Direction (east to West) Is Set up in a Chamber. A Charged Particle Enters the Chamber and Travels Undeflected Along a Straight Path With Constant Speed. What Can You Say About the Initial Velocity of the Particle?

Ans: The initial velocity of the particle could either be parallel or be anti-parallel to the magnetic field. So, it travels along a straight path without undergoing any deflection in the field.

b) A Charged Particle Enters an Environment of a Strong and Non-Uniform Magnetic Field Varying from Point to Point Both in Magnitude and Direction, and Comes Out of it Following a Complicated Trajectory. Would Its Final Speed Equal the Initial Speed If it Suffered No Collisions With the Environment? 

Ans: Yes, the final speed of the charged particle would be equal to its initial speed as the magnetic force has the potential to change the direction of velocity even though not its magnitude.


c) An Electron Travelling West to East Enters a Chamber Having a Uniform Electrostatic Field in the North to South Direction. Specify the Direction in Which a Uniform Magnetic Field Should Be Set up to Prevent the Electron from Deflecting from Its Straight-Line Path.

Ans: An electron travelling from West to East enters a chamber having a uniform electrostatic field along the North-South direction. 

This moving electron stays undeflected when the electric force acting on it is equal and opposite of the magnetic field. 

The magnetic force would stay directed towards the South. Also, according to Fleming’s left-hand rule, the magnetic field must be applied in a vertically downward direction.


19. An Electron Emitted by a Heated Cathode and Accelerated Through a Potential Difference of $2.0kV$, Enters a Region With Uniform Magnetic Field of $0.15T$. Determine the Trajectory of the Electron If the Field 

a) Is Transverse to Its Initial Velocity. 

Ans: We are given, 

Magnetic field strength, $B=0.15T$

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

Mass of the electron, $m=9.1\times {{10}^{-31}}kg$

Potential difference, $V=2.0kV=2\times {{10}^{3}}V$

Now we have the kinetic energy of the electron given by, 

$K.E=eV$

Substituting the given values we get, 

$eV=\frac{1}{2}m{{v}^{2}}$

$\Rightarrow v=\sqrt{\frac{2eV}{m}}$…………………….. (1)

Where, $v$is the velocity of the electron

Since the magnetic force on the electron provides the required centripetal force of the electron, the electron traces a circular path of radius $r$.

Now, the magnetic force on the electron is given by the relation, 

$F=Bev$

Centripetal force, 

${{F}_{C}}=\frac{m{{v}^{2}}}{r}$

$\Rightarrow Bev=\frac{m{{v}^{2}}}{r}$

$\Rightarrow r=\frac{mv}{Be}$………………….. (2)

From the equations (1) and (2), we get, 

$r=\frac{m}{Be}{{\left[ \frac{2eV}{m} \right]}^{\frac{1}{2}}}$

Substituting the given values, 

$\Rightarrow r=\frac{9.1\times {{10}^{-31}}}{0.15\times 1.6\times {{10}^{-19}}}{{\left( \frac{2\times 1.6\times {{10}^{-19}}\times 2\times {{10}^{3}}}{9.1\times {{10}^{-31}}} \right)}^{\frac{1}{2}}}$

$\Rightarrow r=100.55\times {{10}^{-5}}$

$\Rightarrow r=1mm$

Thus, we found that the electron has a circular trajectory of radius 1.0 mm normal to the magnetic field.

b) Makes an Angle of $30{}^\circ $ with the Initial Velocity.

Ans: When the field makes an angle $\theta $ of $30{}^\circ $with initial velocity, the initial velocity will be,

${{v}_{1}}=v\sin \theta $

From equation (2), we can write the following expression:

${{r}_{1}}=\frac{m{{v}_{1}}}{Be}$

$\Rightarrow {{r}_{1}}=\frac{mv\sin \theta }{Be}$

$\Rightarrow {{r}_{1}}=\frac{9.1\times {{10}^{-31}}}{0.15\times 1.6\times {{10}^{-19}}}\left[ \frac{2\times 1.6\times {{10}^{-19}}\times 2\times {{10}^{3}}}{9\times {{10}^{-31}}} \right]\sin 30{}^\circ $

$\Rightarrow r=0.5\times {{10}^{-3}}m=0.5mm$

Clearly, we found that the electron has a helical trajectory of radius$0.5mm$, with axis of the solenoid along the magnetic field direction.


20. A Magnetic Field Set up Using Helmholtz Coils (Described in Exercise $4.16$) is Uniform in a Small Region and Has a Magnitude of $0.75T$. In the Same Region, a Uniform Electrostatic Field Is Maintained in a Direction Normal to the Common Axis of the Coils. A Narrow Beam of (single Species) Charged Particles All Accelerated Through $15kV$ Enters This Region in a Direction Perpendicular to Both the Axis of the Coils and the Electrostatic Field. If the Beam Remains Undeflected When the Electrostatic Field Is $9.0\times {{10}^{-5}}V{{m}^{-1}}$ Make a Simple Guess as to What the Beam Contains. Why Is the Answer Not Unique?

Ans: We are given,

Magnetic field, $B=0.75T$

Accelerating voltage, $V=15kV=15\times {{10}^{3}}V$

Electrostatic field, $E=9\times {{10}^{5}}V{{m}^{-1}}$

Mass of the electron$=m$

Charge of the electron $=e$

Velocity of the electron $=v$

Kinetic energy of the electron $=eV$

Thus, 

$\frac{1}{2}m{{v}^{2}}=eV$

$\Rightarrow \frac{e}{m}=\frac{{{v}^{2}}}{2V}$…………………….. (1)

Since the particle remains undeflected by electric and magnetic fields, we could infer that the electric field is balancing the magnetic field.

$eE=evB$

$\Rightarrow v=\frac{E}{B}$……………………….. (2)

Now we could substitute equation (2) in equation (1) to get,

$\frac{e}{m}=\frac{1}{2}\frac{{{\left( \frac{E}{B} \right)}^{2}}}{V}=\frac{{{E}^{2}}}{2V{{B}^{2}}}$

$\Rightarrow \frac{e}{m}=\frac{{{\left( 9.0\times {{10}^{5}} \right)}^{2}}}{2\times 15000\times {{\left( 0.75 \right)}^{2}}}=4.8\times {{10}^{7}}C/kg$

This value of specific charge $\left( \frac{e}{m} \right)$ is equal to the value of deuteron or deuterium ions. This is not a unique answer. Other possible answers are $H{{e}^{++}}$, $L{{i}^{+++}}$


21. A Straight Horizontal Conducting Rod of Length $0.45m$ and Mass $60g$ is Suspended by Two Vertical Wires at Its Ends. A Current of $5.0A$ is Set up in the Rod Through the Wires.

a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?

Ans: We are given,

Length of the rod, $l=0.45m$

Mass suspended by the wires, $m=60g=60\times {{10}^{-3}}kg$

Acceleration due to gravity, $g=9.8m{{s}^{-2}}$

Current in the rod flowing through the wire, $I=5A$

We could say that magnetic field (B) is equal and opposite to the weight of the wire i.e.,

$BIl=mg$

$\Rightarrow B=\frac{mg}{Il}=\frac{60\times {{10}^{-3}}\times 9.8}{5\times 0.45}$

$\Rightarrow B=0.26T$

Clearly, a horizontal magnetic field of 0.26 T normal to the length of the conductor should be set up.


b) What will be the Total Tension in the Wires If the Direction of Current is Reversed Keeping the Magnetic Field Same as Before? (Ignore the Mass of the Wires.) $g=9.8m{{s}^{-2}}$

Ans: When the direction of the current is reversed, $BIl$ and $mg$ will act downwards. Clearly, the effective tension in the wires is found to be, 

$T=0.26\times 5\times 0.45+\left( 60\times {{10}^{-3}} \right)\times 9.8$

$\Rightarrow T=1.176N$


22. The Wires Which Connect the Battery of an Automobile to Its Starting Motor Carry a Current of $300A$(for a short time). What Is the Force Per Unit Length Between the Wires if they are $70cm$ Long and $1.5cm$ Apart? Is the Force Attractive or Repulsive?

Ans: We are given,

Current in both wires, $I=300A$

Distance between the wires, $r=1.5cm=0.015m$

Length of the two wires, $l=70cm=0.7m$

We know that, Force between the two wires is given by the relation,

$F=\frac{{{\mu }_{0}}{{I}^{2}}}{2\pi r}$

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

As the direction of the current in the wires is found to be opposite, a repulsive force exists between them.


23. A Uniform Magnetic Field of $1.5T$ Exists in a Cylindrical Region of Radius $10.0cm$, its Direction Parallel to the Axis Along East to West. A Wire Carrying Current of $7.0A$ in the North to South Direction Passes Through This Region. What is the Magnitude and Direction of the Force on the Wire If,

a) The Wire Intersects the Axis,

Ans: We are given,

Magnetic field strength, $B=1.5T$

Radius of the cylindrical region, $r=10cm=0.1m$

Current in the wire passing through the cylindrical region, $I=7A$

If the wire intersects the axis, then the length of the wire is the diameter of the cylindrical region. Thus, $l=2r=0.2m$

Angle between magnetic field and current, $\theta =90{}^\circ $

We know that, Magnetic force acting on the wire is given by the relation,

$F=BIl\sin \theta $

$\Rightarrow F=1.5\times 7\times 0.2\times \sin 90{}^\circ $

$\Rightarrow F=2.1N$

Clearly, a force of 2.1 N acts on the wire in a vertically downward direction.

b) The Wire is Turned from N-S to northeast-northwest Direction,

Ans: The new length of the wire after turning it to the northeast-northwest direction can be given as:

${{l}_{1}}=\frac{l}{\sin \theta }$

Angle between magnetic field and current, $\theta =45{}^\circ $

Force on the wire,

$F=BI{{l}_{1}}\sin \theta =BIl=1.5\times 7\times 0.2$

$\Rightarrow F=2.1N$

Thus, a force of 2.1 N acts vertically downward on the wire. This is independent of angle $\theta $ as $l\sin \theta $ is fixed.


c) The Wire in the N-S Direction is Lowered from the Axis by a Distance of $6.0cm$?

Ans: The wire is lowered from the axis by distance, $d=6.0cm$

Let ${{l}_{2}}$be the new length of the wire, 

${{\left( \frac{{{l}_{2}}}{2} \right)}^{2}}=4\left( d+r \right)=4\left( 10+6 \right)=4\times 16$

$\Rightarrow {{l}_{2}}=8\times 2=16cm=0.16m$

Magnetic force that is exerted on the wire is, 

${{F}_{2}}=BI{{l}_{2}}=1.5\times 7\times 0.16$

$\Rightarrow F=1.68N$

Clearly, a force of $1.68N$acts in a vertically downward direction on the wire.

 

24. A Uniform Magnetic Field of $3000G$ is Established Along the Positive Z-Direction. A Rectangular Loop of Sides $10cm$ and $5cm$ Carries a Current of $12A$. What is the Torque on the Loop in the Different Cases Shown in Figure? What Is the Force on Each Case? Which Case Corresponds to Stable Equilibrium?

a)                              


A uniform magnetic established along the positive z-direction with a rectangular loop placed vertically in the yz plane


Ans: We are given, 

Magnetic field strength, $B=3000G=3000\times {{10}^{-4}}T=0.3T$

Length of the rectangular loop, $l=10cm$

Width of the rectangular loop, $b=5cm$

Area of the loop, $A=l\times b=\left( 10\times 5 \right)c{{m}^{2}}=50\times {{10}^{-4}}{{m}^{2}}$

Current in the loop, $I=12A$

Now, we could take the anti-clockwise direction of the current as positive and vice-versa, 

We have the expression for torque given as,

$\vec{\tau }=I\vec{A}\times \vec{B}$

We could see from the given figure that A is normal to the y-z plane and B is directed along the z-axis. Substituting the given values, 

$\tau =12\times \left( 50\times {{10}^{-4}} \right)\hat{i}\times 0.3\hat{k}$

$\Rightarrow \tau =-1.8\times {{10}^{-2}}\hat{j}Nm$

Now, the torque is found to be directed along negative y-direction. Since the external magnetic field is uniform, the force on the loop would be zero. 

b) 


A uniform magnetic established along the positive z-direction with a rectangular loop horizontally placed in the yz plane


Ans: This case is very similar to case (a), and hence, the answer here would be same as (a). 

c) 


A uniform magnetic established along the positive z-direction with a rectangular loop placed in the xz plane


Ans: Torque here would be, 

$\tau =I\vec{A}\times \vec{B}$

$\Rightarrow \tau =-12\left( 50\times {{10}^{-4}} \right)\hat{j}\times 0.3\hat{k}$

$\Rightarrow \tau =-1.8\times {{10}^{-2}}\hat{i}Nm$

The direction here is along the negative x direction and the force is zero. 

d) 


A uniform magnetic established along the positive z-direction with a rectangular loop placed in the xz plane with an angle of 240 degree with x axis


Ans: Torque here would be, 

$\left| \tau  \right|=IAB$

$\Rightarrow \tau =12\times \left( 50\times {{10}^{-4}} \right)\times 0.3$

$\Rightarrow \left| \tau  \right|=1.8\times {{10}^{-2}}Nm$

Here, the direction is found to be at $240{}^\circ $with positive x-direction and the force is zero.

e)


A uniform magnetic established along the positive z-direction with a rectangular loop placed in the xy plane


Ans: Torque, 

$\tau =I\vec{A}\times \vec{B}$

$\Rightarrow \tau =\left( 50\times {{10}^{-4}}\times 12 \right)\hat{k}\times 0.3\hat{k}$

$\Rightarrow \tau =0$

Here, both torque and force are found to be zero. 

f)  

(Image will be Uploaded Soon)

Ans: Torque is given by, 

$\tau =I\vec{A}\times \vec{B}$

$\Rightarrow \tau =\left( 50\times {{10}^{-4}}\times 12 \right)\hat{k}\times 0.3\hat{k}$

$\Rightarrow \tau =0$

Here also both torque and force are found to be zero. 

For the case (e) the direction $I\vec{A}$and $\vec{B}$ is the same and the angle between them is zero. They would come back to equilibrium on being displaced and so the equilibrium is stable. 

For the case (f), the direction of $I\vec{A}$and $\vec{B}$ are opposite and the angle between them is $180{}^\circ $. Here it doesn’t come back to its original position on being disturbed and hence, the equilibrium is unstable. 


25. A Circular Coil of 20 Turns and Radius 10 Cm is Placed in a Uniform Magnetic Field of 0.10 T Normal to the Plane of the Coil. If the Current in the Coil Is 5.0 A, What is the: (The Coil is Made of Copper Wire of Cross-Sectional Area ${{10}^{-5}}{{m}^{2}}$, and the Free Electron Density in Copper Is Given to be about${{10}^{29}}{{m}^{-3}}$).

a) Total Torque on the Coil?

Ans: We are given,

Number of turns on the circular coil, $n=20$

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

Magnetic field strength, $B=0.10T$

Current in the coil, $I=5.0A$

As the angle between force and the normal to the loop is zero, the total torque on the coil is zero. 

So, $\tau =NIAB\sin \theta $is zero.

b) Total Force on the Coil, 

Ans: There is no total force on the coil because the field is uniform.

C) Average Force on Each Electron in the Coil Due to the Magnetic Field?

Ans: Given that,

Cross-sectional area of copper coil, $A={{10}^{-5}}{{m}^{2}}$

Number of free electrons per cubic meter in copper, $N={{10}^{29}}/{{m}^{3}}$

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

Magnetic force, $F=Be{{v}_{d}}$

Where, ${{v}_{d}}$ is drift velocity of electrons given by $\frac{I}{NeA}$

$\Rightarrow F=\frac{BeI}{NeA}=\frac{0.10\times 5.0}{{{10}^{29}}\times {{10}^{-5}}}=5\times {{10}^{-25}}N$

Clearly, the average force on each electron is $5\times {{10}^{-25}}N$.


26. A Solenoid $60cm$ Long and Radius $4.0cm$has 3 Layers of Windings of 300turns Each. A $2.0cm$ Long Wire of Mass $2.5g$ Lies Inside the Solenoid (near Its Centre) Normal to Its Axis; Both the Wire and the Axis of the Solenoid Are in the Horizontal Plane. the Wire Is Connected Through Two Leads Parallel to the Axis of the Solenoid to an External Battery Which Supplies a Current of $6.0a$ in the Wire. What Value of Current (with Appropriate Sense of Circulation) in the Windings of the Solenoid Can Support the Weight of the Wire? $g=9.8m{{s}^{-2}}$

Ans: We are given:

Length of the solenoid, $L=60cm=0.6m$

Radius of the solenoid, $r=4.0cm=0.04m$

It is given that there are 3 layers of windings of 300 turns each.

Total number of turns, $n=3\times 300=900$

Length of the wire, $l=2cm=0.02m$

Mass of the wire, $m=2.5g=2.5\times {{10}^{-3}}kg$

Current flowing through the wire, $i=6A$

Acceleration due to gravity, $g=9.8m{{s}^{-2}}$

Magnetic field produced inside the solenoid, $B=\frac{{{\mu }_{0}}nI}{L}$

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

Current flowing through the windings of the solenoid, $I$

Magnetic force is given by the relation,

$F=Bil=\frac{{{\mu }_{0}}nI}{L}il$

Now, we have the force on the wire equal to the weight of the wire.

$mg=\frac{{{\mu }_{0}}nIil}{L}$

$\Rightarrow I=\frac{mgL}{{{\mu }_{0}}nil}=\frac{2.5\times {{10}^{-3}}\times 9.8\times 0.6}{4\pi \times {{10}^{-7}}\times 900\times 0.02\times 6}$

$\Rightarrow I=108A$

Clearly, the current flowing through the solenoid is 108 A.


27. A Galvanometer Coil Has a Resistance of $12\Omega $and the Metre Shows Full Scale Deflection for a Current of $3mA$. How will you Convert the Metre Into a Voltmeter of Range $0$ to $18V$?

Ans: We are given, 

Resistance of the galvanometer coil, $G=12\Omega $

Current for which there is full scale deflection, ${{I}_{g}}=3mA=3\times {{10}^{-3}}A$

Range of the voltmeter needs to be converted to $18V$.

Let a resistor of resistance R be connected in series with the galvanometer to convert it into a voltmeter. This resistance can be given as:

$R=\frac{V}{{{I}_{g}}}-G$

Substituting the given values we get, 

$R=\frac{18}{3\times {{10}^{-3}}}-12=6000-12$

$\Rightarrow R=5988\Omega $

Clearly, we found that a resistor of resistance $5988\Omega $is to be connected in series with the given galvanometer. 


28. A Galvanometer Coil Has a Resistance of $15\Omega $ and the Metre Shows Full Scale Deflection for a Current of $4mA$. How will you Convert the Metre Into an Ammeter of Range $0$ to $6A$?

Ans: We are given,

Resistance of the galvanometer coil, $G=15\Omega $

Current for which the galvanometer shows full scale deflection, ${{I}_{g}}=4mA=4\times {{10}^{-3}}A$

We said that, Range of the ammeter needs to be $6A$.

In order to convert the given galvanometer into an ammeter, a shunt resistor of resistance S is to be connected in parallel with the galvanometer. 

The value of S could be given as:

$S=\frac{{{I}_{g}}G}{I-{{I}_{g}}}$

Substituting the given values we get, 

$S=\frac{4\times {{10}^{-3}}\times 15}{6-4\times {{10}^{-3}}}=\frac{0.06}{5.996}\approx 0.01\Omega $

$\Rightarrow S=10m\Omega $

Clearly, we found that a \[10m\Omega \] shunt resistor is to be connected in parallel with the galvanometer.

NCERT Solutions for Class 12 Physics Chapter 4 PDF

Students preparing for their class 12 boards must focus on the fourth chapter, which is crucial for the exam. Moreover, it is also significant for other competitive exams. There are multiple subtopics in chapter 4 Physics class 12, and they demand a thorough understanding. While studying this topic, students should also refer to a solutions book for learning these topics quickly.


These solutions are readily available online in PDF formats. Several dedicated E-learning platforms offer these solutions free of cost. Students can easily download these and study it as per their convenience. The PDFs of chapter 4 Physics class 12 NCERT solutions are curated by experts with years of experience. Thus, these solutions are accurate and highly reliable.


Magnetism is caused by moving charges or the flow of charge. Our revision notes on Moving Charges and Magnetism elaborate on the fact that magnetic fields exert further forces on the flow of charge. This, in turn, exerts a force on other magnets as well. This phenomenon occurs due to the presence of constant moving charges. For the benefits of these NCERT Solutions students should consider referring to the PDF for class 12 Physics chapter 4. This will enable them to acquire in-depth knowledge about these topics and thereby will help them to score well.


Chapter 4 Physics Class 12: Subtopics and Exercises

Moving charges and magnetism is divided into eleven subtopics, each of which is important for CBSE class 12 as well as JEE and NEET. To do well on any of these tests, students must focus on gaining in-depth knowledge in all of these areas. They may use NCERT solutions for class 12 Physics chapter 4 to help them in this procedure. These solutions have been specifically created to make this subject more comprehensible. Also, it will make it easier for them to complete the workouts.


There are approximately twenty-eight questions in the exercise of this chapter. Mostly all the questions are based on calculations. To solve these questions, students must have a clear concept of all the sections of the chapter.


Students who face difficulty in solving these sums or the other questions can consider referring to NCERT solutions for class 12 Physics Chapter 4 Moving Charges and Magnetism. As most of the questions are mathematical, everyone should cross-check it from the solutions book. This will help them to identify their mistakes and rectify them.


Topics Covered in Ch 4 Physics Class 12 NCERT Solutions

The answers given in the solutions are elaborately explained. The calculations are shown step-by-step which will be beneficial for the students to understand the logic behind each step. Moreover, students studying NCERT solutions for class 12 Physics Moving Charges and Magnetism will get a clear idea about the following:

  • Magnetic force and its sources.

  • Magnetic field (shown in the image below).


(image will be uploaded soon)


  • Lorentz Field

  • Magnetic force

  • Motion in a magnetic field

  • Velocity Selector

  • Cyclotron

  • Biot-Savart Law

  • Ampere’s Circuital Law

  • The Solenoid

  • The Toroid

  • Magnetic dipole

  • The Ampere

  • The moving coil galvanometer

Moving charges and Magnetism has a share of 8 marks in the exam. The distribution of marks are as follows:


Moving Charges and Magnetism Class 12: Marks Distribution


Very Short Answer (1 Mark)

Short Answer II  (2 Mark)

Long Answer (5 Mark)

Total Marks

1

1

1

8

 

Benefits of NCERT Solutions for Class 12 Physics Chapter 4

To ensure good grades in Physics, students should make sure that their concept regarding every topic is clear. However, studying only from the textbooks is not sufficient to achieve this, they must refer to class 12 NCERT solutions Physics ch 4. Learning from the solutions has many benefits, some of which are as follows:

  • It is written in an easy language.

  • The solutions are designed by faculty having years of knowledge; thereby it is accurate.

  • It has illustrations which make learning easier.

  • The calculations are explained in detailed steps.


Conclusion 

NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges and Magnetism offer comprehensive and invaluable learning resources for students. By addressing key concepts such as Lorentz force, Biot-Savart law, and Ampere's circuital law, these solutions enable a deeper understanding of the intricate relationship between moving charges and magnetism. The step-by-step explanations and solved examples provided in these solutions aid students in tackling complex problems with confidence. Additionally, the emphasis on real-life applications fosters an appreciation for the practical relevance of the subject. Overall, NCERT Solutions for Class 12 Physics Chapter 4 empower learners to grasp the fundamentals and excel in their academic journey.

FAQs on NCERT Solutions for Class 12 Physics Chapter 4 - Moving Charges And Magnetism

1. How Should I Prepare Magnetism and Moving Charges for my Boards?

While preparing this chapter, students should first go through the previous year papers. After finding the most relevant questions, they should start their preparation. Moreover, students should emphasise on practising sums as these have a majority of share in the question paper. They should revise the formulas regularly.


The questions are designed to check the analytical and quantitative aptitudes of the students. Therefore, they should focus on understanding the concepts well. To do this, they can take the help of Moving Charges and Magnetism class 12 NCERT solutions. This will help them to gain comprehensive knowledge about the topics. Besides, students while solving sums can refer to these solutions to identify their mistakes and rectify them.

2. Why Should I Refer to the NCERT Solution of Physics Class 12 Chapter 4?

Physics demands an in-depth understanding of the concepts. Without understanding the topics, students will not be able to score well in their exams. Moreover, this chapter helps students to prepare for other competitive exams like JEE and NEET as well.


Considering the importance of this chapter, students should study it thoroughly. NCERT solutions are designed to help them in the process. All the answers are explained elaborately, which allows students to understand the logic behind it. Besides, these are curated by an experienced faculty, and therefore the answers are highly reliable. The solutions strictly adhere to the CBSE pattern and help students to score well in their exams.

3. What is a Magnetic Field?

Magnetic field refers to an area around a magnet which shows its magnetic force. The magnetic field has direction as well as magnitude. This is an essential topic of class 12 Physics. Students while studying this should refer to Physics class 12, chapter 4, NCERT solutions. The solutions will help them to understand this topic better. While referring to the guide, they will also come across other related concepts like a magnetic force, motion in a magnetic field, velocity selector and others. After understanding these concepts, students will be able to solve the questions quickly.

4. Why should I choose Vedantu’s NCERT Solutions as my first option for preparation?

Our science masters have rendered these solutions in an easily understandable letter. Methodological flow has been followed in this chapter to prepare yourself for the exam. You can use the techniques given for topics and subtopics of this chapter to score more marks. These solutions will assist you in attaining a better understanding of various basic concepts composed in this chapter with expertise. You can score well if you have secured 100% confidence to answer any question asked from this chapter. Our faculty made sure that CBSE and NCERT guidelines are strictly followed while drafting these solutions.

5. What are the key features of NCERT Solutions for Class 12 Physics Chapter 4?

The experienced team at Vedantu has diligently prepared all of the questions in NCERT Solutions for Class 12 Physics Chapter 4 Moving Charge and Magnetism, and the major benefits of these solutions are:

  • These solutions are provided free of cost. Thus, anyone can access them.

  • The solutions are provided in a PDF format that can be downloaded and printed for revision.

  • The professional team of Vedantu makes sure to structure all the solutions based only on the CBSE curriculum.

6. Is it necessary to use the NCERT Solutions for Class 12 Physics Chapter 4 PDF?

Class 12 Physics is important not only for the board exams but also for other competitive exams such as JEE, NEET, etc. Thus, one should clear all the concepts to score good grades in CBSE boards and other competitive exams. In order to accomplish this, it is advisable to refer to NCERT Books and NCERT Solutions for a comprehensive understanding of the concepts. The solutions to each question are thoroughly discussed on Vedantu.

7. What is Permittivity and Permeability Class 12 Physics?

A physical property that specifies how an electric field influences and is affected by a medium is known as electric permittivity. It is determined by a material's ability to polarise in response to an applied field. Similarly, magnetic permeability is an ability of a substance to acquire magnetisation in magnetic fields. It is a measure of how far a magnetic field can penetrate matter.

8. What are the concepts explained in Class 12 Physics Chapter 4?

Following are the concepts that are explained in Class 12 Physics Chapter 4:

  • Magnetic Force

  • Sources and Fields

  • Magnetic field and Lorentz force

  • Motion In Magnetic Field

  • Motion In Combined Electric and Magnetic Fields

  • Velocity Selector

  • Cyclotron

  • Biot-Savart Law

  • Ampere’s Circuital Law

  • Solenoid and Toroid

  • Torque On Current Loop, Magnetic Dipole

  • Circular Current Loop as A Magnetic Dipole

  • The Magnetic Dipole Moment of a Revolving Electron

  • The Moving Coil Galvanometer

9. Where can I find PDFs for NCERT Solutions Class 12 Physics?

Ans: You can find PDFs of NCERT Solutions Class 12 Physics on Vedantu. You can check out the app or the website for the same. Vedantu provides students with well-curated NCERT Solutions prepared by the Subject-Matter Experts and are totally based on the curriculum presented by CBSE. This free and downloadable study material will not only help to score good grades in Class 12 Boards but also in other competitive exams. Students can also access the study material from Vedantu’s App. All the resources are available free of cost.