NCERT Solutions Class 12 Physics Chapter 4 Moving Charges and Magnetism

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, $r=2.5m$

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.

Moving Charges and Magnetism Chapter Summary - 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.

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.

Magnetic Field at Centre O in Different Conditions 

Condition	Figure	Magnetic Field
Are subtends angle at $\theta$ the centre		$B= \frac{\mu _{0}}{4\pi }\cdot \frac{\theta _{i}}{r}$
Are subtends angle $(2^{\pi }-^{\theta })$ at the centre		$B=\frac{\mu _{0}}{4\pi }\times\frac{ (2\pi-\theta )i}{r}$
Semi – circular are		$B= \frac{\mu _{0}}{4\pi }\cdot \frac{\pi i}{r}=\frac{\mu _{0}i}{4r}$
Three quarter semi  circular current carrying are		$B= \frac{\mu _{0}}{4\pi }\cdot \frac{(2\pi -\frac{\pi }{2})i}{r}$
Circular current carrying are		$B= \frac{\mu _{0}}{4\pi }\cdot \frac{2\pi i}{r}=\frac{\mu _{0}i}{2r}$
Concentric co – planer circular loops carries current in the same direction		$B= \frac{\mu _{0}}{4\pi }2\pi(\frac{1}{r_{1}}+\frac{1}{r{2}})$
Concentric co-planer circular loops carries current in the opposite direction		$B_{2}= \frac{\mu _{0}}{4\pi }2\pi i\left [ \frac{1}{r_{1}} -\frac{1}{r_{2}}\right ]$
Concentric loops but their planes are perpendicular to each other		$B=\sqrt{B_{1}^{2}+B_{2}^{2}}$ $=\frac{_{\mu _{0}}}{2r}\sqrt{i_{1}^{2}+i_{2}^{2}}$
Concentric loops but their planes are at an angle $\theta$ with each other		$B=\sqrt{B_{1}^{2}+B_{2}^{2}+2B_{1}B_{2}}$
Distribution of current across the diameter		B = 0
Distribution of current between any two points on the circumference		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.

• 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.

Overview of Deleted Syllabus for CBSE Class 12 Physics Moving Charges and Magnetism

Conclusion

NCERT Class 12 Physics Chapter 4 Exercise Solutions on  Moving Charges and Magnetism provided by Vedantu provides a thorough exploration of the relationship between electricity and magnetism. This chapter lays the foundation for understanding how moving charges generate magnetic fields and how these fields exert forces on other moving charges and currents. The chapter equips students with the theoretical and practical knowledge needed within the fields of electrical and electronic engineering. In previous year's question papers, typically around 3–4 questions were asked from this chapter.

Other Study Material for CBSE Class 12 Physics Chapter 4

Chapter-Specific NCERT Solutions for Class 12 Physics

Given below are the chapter-wise NCERT Solutions for Class 12 Physics. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.