Chapter 4 Moving Charges and Magnetism

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

NCERT Solutions for Class 12 Physics Chapter 4 (Moving Charges and Magnetism) play a crucial role in helping students build a strong conceptual foundation for both CBSE board exams and competitive exams like JEE Main, JEE Advanced, and NEET. This chapter will give students the mathematical foundations for finding magnetic forces and fields using the Biot-Savart Law, and Ampere's Circuital Law. Understanding these concepts is essential for students as they lead to many of the systems that utilitise this technology, including electric motors and particle accelerators.

The NCERT Solutions for Class 12 Physics Chapter 4 (Moving Charges and Magnetism) will help students to learn how to use both the "Right-Hand Rules" and vector cross-products to calculate the magnetic interaction between moving charges and others around them. Well-explained NCERT solutions enable students to strengthen fundamentals, practice important questions, and perform better in term exams, board exams, and national-level competitive tests.

1.0Class 12 Physics Chapter 4 Moving Charges and Magnetism: Key Concepts

Class 12 Physics Chapter 4, Moving Charges and Magnetism, explains the magnetic effects of electric current and the force acting on charged particles moving through a magnetic field. The main concepts covered in this chapter include:

Magnetic Force and Lorentz Force: Understand the force acting on a charged particle moving in a magnetic field and the combined effect of electric and magnetic fields.
Biot-Savart Law: Learn how to calculate the magnetic field produced by a current-carrying conductor using the Biot-Savart law.
Magnetic Field Due to a Straight Wire and Circular Loop: Find the magnetic field produced by a straight current-carrying wire and a circular loop, including the magnetic field along the axis of a circular coil.
Ampere’s Circuital Law: Use Ampere’s circuital law to calculate magnetic fields in cases with suitable symmetry, such as solenoids and toroids.
Force Between Two Parallel Currents: Understand the force between two parallel current-carrying conductors and why parallel currents attract while opposite currents repel.
Torque on a Current Loop: Learn how a magnetic field produces torque on a current-carrying loop and understand its role in the working of electric motors and galvanometers.
Moving Coil Galvanometer: Study the construction and working of a moving coil galvanometer and learn how it can be converted into an ammeter and a voltmeter.

2.0NCERT Solutions Class 12 Physics Chapter 4 Moving Charges and Magnetism : Detailed Solutions

SOLVED EXAMPLES

A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field $B$ (shown in fig.). What is the magnitude of the magnetic field?

Conducting rod carrying current I in a uniform magnetic field, with its length l and downward gravitational force mg shown.

Sol. From Lorentz formula, we find that there is an upward force F, of magnitude I/B For mid-air suspension, this must be balanced by the force due to gravity :

$mg = I / B B = \frac{mg}{I l} = \frac{0.2 \times 9.8}{2 \times 1.5} = 0.65 T$

Note that it would have been sufficient to specify $m / l$ , the mass per unit length of the wire. The earth's magnetic field is approximately $4 \times 1 0^{- 5} T$ and we have ignored it.

If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig.), which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge).

Coordinate axes showing a velocity vector along the y-direction in a uniform magnetic field directed into the plane.

Sol. The velocity $v$ of particle is along the x-axis, while $B$ , the magnetic field is along the y-axis, so $v \times B$ is along the z-axis (screw rule or right-hand thumb rule). So, (a) for electron it will be along -z axis. (b) for a positive charge (proton) the force is along +z axis. (Both directions are considered for given instant)

What is the radius of the path of an electron (mass $9 \times 1 0^{- 31} kg$ and charge $1.6 \times 1 0^{- 19} C$ ) moving at a speed of $3 \times 1 0^{7} m / s$ in a magnetic field of $6 \times 1 0^{- 4} T$ perpendicular to it? What is its frequency? Calculate its energy in keV. $(1 eV = 1.6 \times 1 0^{- 19} J)$ Sol. Using equation we find, $r = mv / (qB)$

$r = 9 \times 1 0^{- 31} kg \times 3 \times 1 0^{7} ms^{- 1} / (1.6 \times 1 0^{- 19} C \times 6 \times 1 0^{- 4} T) r = 28 \times 1 0^{- 2} m = 28 cm v = v / (2 π r) = \frac{3 \times 1 0 ^{7}}{2 \times 3.14 \times 28 \times 1 0 ^{- 2}} = 17 \times 1 0^{6} s^{- 1} v = 17 \times 1 0^{6} Hz = 17 MHz E = (1/2) mv^{2} = (1/2) 9 \times 1 0^{- 31} kg \times 9 \times 1 0^{14} m^{2} / s^{2} = 40.5 \times 1 0^{- 17} J E \approx 4 \times 1 0^{- 16} J = 2.5 keV .$

An element $Δ l = Δ x \hat{i}$ is placed at the origin and carries a large current $I = 10 A$ (figure). What is the magnetic field on the y-axis at a distance of 0.5 m. $Δ x = 1 cm$ .

Charged particle moving along the x-axis near point P on the y-axis, with the separation and displacement shown.

Sol. $∣ d B ∣ = \frac{μ _{0}}{4 π} \frac{Id l s i n θ}{r ^{2}}$ [From Biot-Savart Law]

$d l = Δ x = 1 0^{- 2} m, I = 10 A, r = 0.5 m = y, μ_{0} /4 π = 1 0^{- 7} \frac{Tm}{A} & θ = 9 0^{\circ}; sin θ = 1 ∣ d B ∣ = \frac{1 0 ^{- 7} \times 10 \times 1 0 ^{- 2}}{25 \times 1 0 ^{- 2}} = 4 \times 1 0^{- 8} T$

The direction of the field is in the +z-direction. This is so since,

$d l \times r = Δ x \hat{i} \times y \hat{j} = y Δ x (\hat{i} \times \hat{j}) = y Δ x \hat{k}$

We remind you of the following cyclic property of cross-products.

$\hat{i} \times \hat{j} = \hat{k}; \hat{j} \times \hat{k} = \hat{i}; \hat{k} \times \hat{i} = \hat{j}$

Note that the field is small in magnitude.

A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0 cm as shown in Fig. (a). Consider the magnetic field B at the centre of the arc. (a) What is the magnetic field due to the straight segments? (b) In what way the contribution to B from the semicircle differs from that of a circular loop and in what way does it resemble? (c) Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Fig. (b)?

Two semicircular current paths showing opposite bending directions, with current arrows along the paths.

Sol.

(a) $d ℓ$ and $r$ for each element of the straight segments are parallel. Therefore, $d ℓ \times r = 0$ . Straight segments do not contribute to $∣ B ∣$ . (b) For all segments of the semi-circular arc. $d ℓ \times r$ are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semi-circular arc is given by the right-hand rule and magnitude is half that of a circular loop. So, magnitude of B is

$B = \frac{μ _{0} I}{4 R} = \frac{4 π \times 1 0 ^{- 7} \times 12}{4 \times 2 \times 1 0 ^{- 2}} = 6 \times 3.14 \times 1 0^{- 5} T B = 1.9 \times 1 0^{- 4} T$

Thus B is $1.9 \times 1 0^{- 4} T$ normal to the plane of the paper going into it. (c) Same magnitude of B but opposite in direction to that in (b).

Consider a tightly wound 100 turn coil of radius 10 cm. carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil ? Sol. Since the coil is tightly wound, we may take each circular element to have the same radius $R = 10 cm = 0.1 m$ . The number of turns $N = 100$ . The magnitude of the magnetic field is,

$B = \frac{μ _{0} NI}{2 R} = \frac{4 π \times 1 0 ^{- 7} \times 1 0 ^{2} \times 1}{2 \times 1 0 ^{- 1}} B = 2 π \times 1 0^{- 4} = 6.28 \times 1 0^{- 4} T$

Figure shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region $r < a$ and $r > a$ .

Circular Amperian loops around a current-carrying conductor, showing inner and outer paths used to determine magnetic field.

Sol.

(a) Consider the case $r > a$ . The Amperian loop, labelled 2, is a circle concentric with the cross-section. For this loop.

$L = 2 π r I_{e} = Current enclosed by the loop = I$

The result is the familiar expression for a long straight wire By using Ampere's law

$B (2 π r) = μ_{0} I B_{out} = \frac{μ _{0} I}{2 π r} B_{out} \propto \frac{1}{r} (r > a)$

(b) when $r < a$ Now the current enclosed $I_{e}$ is not I, but is less than this value. Since the current distribution is uniform, the current enclosed is,

$I_{e} = I (\frac{π r _{1}^{2}}{π a ^{2}}) = \frac{I r _{1}^{2}}{a ^{2}}$

Using Ampere's law, $B (2 π r_{1}) = μ_{0} \frac{Ir _{1}^{2}}{a ^{2}}$

$B_{in} = (\frac{μ _{0} I}{2 π a ^{2}}) r_{1} B_{in} \propto (r_{1} = r < a)$

Graph showing magnetic field variation with radial distance inside and outside a current-carrying cylindrical conductor.

Figure shows a plot of the magnitude of B with distance r from the centre of the wire. The direction of the field is tangential to the respective circular loop (1 or 2) and given by the right-hand rule described earlier in this section. This example possesses the required symmetry so that Ampere's law can be applied readily.

A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5A. What is the magnitude of the magnetic field inside the solenoid ? Sol. The number of turns per unit length is,

$n = \frac{500}{0.5} = 1000 turns / m$

The length $l = 0.5 m$ and radius $r = 0.01 m$ . Thus, $l / a = 50$ i.e., $l ≫ a$ . Hence, we can use the long solenoid formula, i.e.

$B = μ_{0} nI B = 4 π \times 1 0^{- 7} \times 1 0^{3} \times 5 B = 6.28 \times 1 0^{- 3} T$

The horizontal component of the earth's magnetic field at a certain place is $3.0 \times 1 0^{- 5} T$ and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west; (b) south to north? Sol. $F = I ℓ \times B$

$F = I / B sin θ [Magnitude of force]$

The force per unit length is

$f = F / l = IB sin θ$

(a) When the current is flowing from east to west.

$θ = 9 0^{\circ}$

Hence,

$f = IB f = 1 \times 3 \times 1 0^{- 5} = 3 \times 1 0^{- 5} Nm^{- 1}$

This is larger than the value $2 \times 1 0^{- 7} Nm^{- 1}$ quoted in the definition of the ampere. Hence it is important to eliminate the effect of the earth's magnetic field and other stray field while standardising the ampere. The direction of the force is downwards. This direction may be obtained by the directional property of cross product of vectors.

(b) When the current is flowing from south to north.

$θ = 0^{\circ}, f = 0$

Hence there is no force on the conductor.

A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A. (a) What is the field at the centre of the coil ? (b) What is the magnetic moment of this coil ? The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 2T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of 90° under the influence of the magnetic field. (c) What are the magnitudes of the torques on the coil in the initial and final position? (d) What is the angular speed acquired by the coil when it has rotated by 90°? The moment of inertia of the coil is $0.1 kg m^{2}$ .

Sol.

(a) The magnetic field at the centre of the coil is

$B = \frac{μ _{0} NI}{2 R}$

Here,

$N = 100; I = 3.2 A and R = 0.1 m . Hence B = \frac{4 π \times 1 0 ^{- 7} \times 100 \times 3.2}{2 \times 1 0 ^{- 1}} = \frac{4 \times 1 0 ^{- 5} \times 10}{2 \times 1 0 ^{- 1}} B = 2 \times 1 0^{- 3} T (using π \times 3.2 = 10)$

The direction is given by the right-hand thumb rule. (b) The magnetic moment is given by below eq.,

$m = N IA = NI π r^{2} m = 100 \times 3.2 \times 3.14 \times 1 0^{- 2} = 10 A m^{2}$

The direction is once again given by the right-hand thumb rule. (c) $τ = ∣ m \times B ∣$

$τ = mB sin θ$

Initially, $θ = 0$ . Thus, initial torque $τ_{i} = 0$ . Finally, $θ = π /2$ (or 90°). Thus, final torque $τ_{f} = mB = 10 \times 2 = 20 Nm$ . (d) From Newton's second law,

$ϑ \frac{d ω}{dt} = mB sin θ$

where $ϑ$ is the moment of inertia of the coil. From chain rule,

$\frac{d ω}{dt} = \frac{d ω}{d θ} \frac{d θ}{dt} = \frac{d ω}{d θ} ω$

Using this,

$ϑ ω d ω = mB sin θ d θ$

Integrating from $θ = 0$ to $θ = π /2$ .

$ϑ \int_{0}^{ω_{f}} ω d ω = mB \int_{0}^{π /2} sin θ d θ ϑ \frac{ω _{f}^{2}}{2} = - mB [cos θ]_{0}^{π /2} = mB ω_{f} = (\frac{2 mB}{ϑ})^{1/2} = (\frac{2 \times 20}{1 0 ^{- 1}})^{1/2} = 20 rad s^{- 1} .$

(a) A current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e., turns about the vertical axis). (b) A current-carrying circular loop is located in a uniform external magnetic field. If the loop is free to turn, what is its orientation of stable equilibrium? Show that in this orientation, the flux of the total field (external field + field produced by the loop) is maximum. (c) A loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible, why does it change to a circular shape?

Sol.

(a) No, because that would require $τ$ to be in the vertical direction. But $τ = I A \times B$ , and since $A$ of the horizontal loop is in the vertical direction, $τ$ would be in the plane of the loop for any $B$ . (b) Orientation of stable equilibrium is one where the area vector A of the loop is in the direction of external magnetic field. In this orientation, the magnetic field produced by the loop is in the same direction as external field, both normal to the plane of the loop, thus giving rise to maximum flux of the total field. (c) It assumes circular shape with its plane normal to the field to maximise flux, since for a given perimeter, a circle encloses greater area than any other shape.

In the given circuit (Fig.) the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance $R_{G} = 60.00Ω$ ; (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance $r_{s} = 0.02Ω$ ; (c) is an ideal ammeter with zero resistance?

Simple electric circuit showing a 3 Ω resistor, ammeter, and 3 V battery connected in s

Sol.

(a) Total resistance in the circuit is, $R_{G} + 3 = 63Ω$ . Hence $I = 3/63 = 0.048 A$ (b) Resistance of the galvanometer converted to an ammeter is,

$\frac{R _{G} r _{s}}{R _{G} + r _{s}} = \frac{60Ω \times 0.02Ω}{( 60 + 0.02 ) Ω} ≃ 0.02Ω$

Total resistance in the circuit is

$0.02Ω + 3Ω = 3.02Ω. Hence, I = 3/3.02 = 0.99 A$

$I = 3/3 = 1.00 A$

EXERCISE QUESTIONS WITH SOLUTIONS

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A . What is the magnitude of the magnetic field B at the centre of the coil?

Sol. $N = 100, r = 8.0 cm = 0.08 m$ & $I = 0.4 A$ Magnitude of the magnetic field at the centre of the coil,

$B \Rightarrow B B = \frac{μ _{0} NI}{2 r} = \frac{4 π \times 1 0 ^{- 7} \times 100 \times 0.4}{2 \times 0.08} = 3.14 \times 1 0^{- 4} T$

A long straight wire carries a current of 35A. What is the magnitude of the field B at a point 20 cm from the wire?

Sol. $I = 35 A & r = 20 cm = 0.2 m$ Magnitude of the magnetic field at this point,

$B \Rightarrow B B = \frac{μ _{0} I}{2 π r} = \frac{4 π \times 1 0 ^{- 7} \times 35}{2 π \times 0.2} = 3.5 \times 1 0^{- 5} T$

A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire. Sol. $I = 50 A, r = 2.5 m$

$B B = \frac{μ _{0} I}{2 π r} \Rightarrow B = \frac{4 π \times 1 0 ^{- 7} \times 50}{2 π \times 2.5} = 4 \times 1 0^{- 6} T$

Magnetic field line around a current-carrying straight conductor, showing current direction and magnetic field at point E.

According to the Maxwell's right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line? Sol. Current in the power line $I = 90 A$ , Point is located below the power line at distance, $r = 1.5 m$ , Hence, magnetic field at that point,

$B B = \frac{μ _{0} I}{2 π r} \Rightarrow B = \frac{4 π \times 1 0 ^{- 7} \times 90}{2 π \times 1.5} = 1.2 \times 1 0^{- 5} T$

Diagram showing a power line carrying current and a point P below it in a vertical plane, with the magnetic field direction indicated.

The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell's right hand thumb rule, the direction of the magnetic field is towards the South.

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T? Sol. $I = 8 A, B = 0.15 T, θ = 3 0^{\circ}$ . Magnetic force per unit length on the wire

$\Rightarrow \frac{F}{ℓ} = BI sin θ \frac{F}{ℓ} = 0.15 \times 8 \times sin 3 0^{\circ} = 0.6 N / m$

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire? Sol. $ℓ = 3 cm = 0.03 m, I = 10 A, B = 0.27 T, θ = 9 0^{\circ}$ Magnetic force exerted on the wire, $F = BI ℓ sin θ$

$F = 0.27 \times 10 \times 0.03 \times sin 9 0^{\circ} = 8.1 \times 1 0^{- 2} N$

The direction of the force can be obtained from Fleming's left hand rule.

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A. Sol. $I_{A} = 8.0 A, I_{B} = 5.0 A, r = 4.0 cm = 0.04 m$ Length of a section of wire A,

$L = 10 cm = 0.1 m$

Force exerted on length L due to the magnetic field, $F = \frac{μ _{0} I _{A} I _{B} L}{2 π r}$

$F = \frac{4 π \times 1 0 ^{- 7} \times 8 \times 5 \times 0.1}{2 π \times 0.04} = 2 \times 1 0^{- 5} N$

This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A. estimate the magnitude of B inside the solenoid near its centre. Sol. Length of the solenoid $ℓ = 80 cm = 0.8 m$ , There are five layers of windings of 400 turns each on the solenoid. ∴ Total number of turns on the solenoid,

$N = 5 \times 400 = 2000$

Diameter of the solenoid,

$D = 1.8 cm = 0.018 m$

Current carried by the solenoid, $I = 8.0 A$ Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,

$B B = \frac{μ _{0} NI}{l} = \frac{4 π \times 1 0 ^{- 7} \times 2000 \times 8}{0.8} = 2.5 \times 1 0^{- 2} T$

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80T. What is the magnitude of torque experienced by the coil? Sol. $ℓ = 10 cm = 0.1 m, I = 12 A, N = 20, B = 0.80 T$ Area of the square coil

$A = ℓ^{2} = 0.1 \times 0.1 = 0.01 m^{2}$

Angle made by the plane of the coil with magnetic field, $θ = 3 0^{\circ}$ Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation.

Magnetic dipole torque calculation showing the angle between the area vector and magnetic field.

$τ = 0.96 N - m$

Two moving coil galvanometers, $M_{1}$ and $M_{2}$ have the following particulars:

$R_{1} = 10Ω, N_{1} = 30, A_{1} = 3.6 \times 1 0^{- 3} m^{2}, B_{1} = 0.25 T, R_{2} = 14Ω, N_{2} = 42, A_{2} = 1.8 \times 1 0^{- 3} m^{2}, B_{2} = 0.50 T$

(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of $M_{2}$ and $M_{1}$ . Sol. When current carrying coil is placed in magnetic field then torque on coil,

$τ = C θ = BI NA (for M ⊥ B)$

Current sensitivity (C.S.) $= θ / I$

$= \frac{BNA}{C} \propto BNA (C is same for M_{1} and M_{2}) \frac{( C . S . ) _{2}}{( C.S ) _{1}} = \frac{B _{2} N _{2} A _{2}}{B _{1} N _{1} A _{1}} = \frac{0.5 \times 42 \times 1.8 \times 1 0 ^{- 3}}{0.25 \times 30 \times 3.6 \times 1 0 ^{- 3}} = 1.4$

Voltage sensitivity (V.S.) $= θ / V = \frac{θ}{IR} = \frac{C.S.}{R}$

$\frac{( V . S . ) _{2}}{( V . S . ) _{1}} = \frac{( C.S. ) _{2}}{( C.S. ) _{1}} \times \frac{R _{1}}{R _{2}} = 1.4 \times \frac{10}{14} = 1$

In a chamber, a uniform magnetic field of 6.5 $G (1 G = 1 0^{- 4} T)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 1 0^{6} ms^{- 1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.

$(e = 1.6 \times 1 0^{- 19} C, m_{e} = 9.1 \times 1 0^{- 31} kg)$

Sol. $B = 6.5 G = 6.5 \times 1 0^{- 4} T, v = 4.8 \times 1 0^{6} m s^{- 1}$ , When electron projected perpendicular to the uniform magnetic field then maximum magnetic force of constant magnitude acts on electron.

$F_{m} F_{m} = evB = q (v \times B), (∵ v ⊥ B) = constant [Magnitude of force]$

The direction of magnetic force is always perpendicular to the direction of motion of electron $(F_{m} ⊥ v)$ , so magnetic force acts as a centripetal force. Hence electron describe uniform circular motion. Radius of circular orbit -

$evB = \frac{mv ^{2}}{r} r = \frac{mv}{eB} = \frac{9.1 \times 1 0 ^{- 31} \times 4.8 \times 1 0 ^{6}}{1.6 \times 1 0 ^{- 19} \times 6.5 \times 1 0 ^{- 4}} r = 0.042 m = 4.2 cm$

In Q. 11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain. Sol. Frequency of revolution of electron :

$f = \frac{eB}{2 π m} = \frac{1.6 \times 1 0 ^{- 19} \times 6.5 \times 1 0 ^{- 4}}{2 \times 3.14 \times 9.1 \times 1 0 ^{- 31}} f = 18 \times 1 0^{6} rev / sec$

For uniform circular motion frequency of electron does not depends on speed of electron.

(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of $6 0^{o}$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (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.) Sol. (a) Magnetic torque on coil :-

$τ τ τ = BINA sin θ = 1 \times 6 \times 30 \times 3.14 \times (8 \times 1 0^{- 2})^{2} \times sin 6 0^{\circ} = 3.1 N - m = Mechanical torque on coil = Magnetic torque on coil = 3.1 N - m$

(b) No, the answer is not changed because torque on coil is does not depends on shape of coil i.e. two different shapes have same geometrical areas.

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Physics chapter-wise, covering exercise answers, key formulas, and step-by-step solutions to improve concept understanding and question practice.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 4

Correct Direction of Force and Field: The solutions explain the Right-Hand Thumb Rule and Fleming’s Left-Hand Rule clearly to help students find the correct direction of magnetic field and force.
Step-by-Step Derivations: Important derivations, such as the magnetic field on the axis of a circular loop and the force between two parallel current-carrying wires, are explained in simple steps.
Clear Understanding of Solenoid and Toroid: The solutions explain the magnetic field inside and outside a solenoid and the magnetic field of a toroid using Ampere’s circuital law.
Galvanometer Conversion: Step-by-step methods are provided to calculate the resistance needed to convert a galvanometer into an ammeter or voltmeter.
Use of Vector Products: The solutions explain how to use formulas such as $q (v \times B) and I (l \times B)$ to solve questions involving the direction of force and magnetic field.

Frequently Asked Questions

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

1.0Class 12 Physics Chapter 4 Moving Charges and Magnetism: Key Concepts

Magnetic Force and Lorentz Force: Understand the force acting on a charged particle moving in a magnetic field and the combined effect of electric and magnetic fields.
Biot-Savart Law: Learn how to calculate the magnetic field produced by a current-carrying conductor using the Biot-Savart law.
Magnetic Field Due to a Straight Wire and Circular Loop: Find the magnetic field produced by a straight current-carrying wire and a circular loop, including the magnetic field along the axis of a circular coil.
Ampere’s Circuital Law: Use Ampere’s circuital law to calculate magnetic fields in cases with suitable symmetry, such as solenoids and toroids.
Force Between Two Parallel Currents: Understand the force between two parallel current-carrying conductors and why parallel currents attract while opposite currents repel.
Torque on a Current Loop: Learn how a magnetic field produces torque on a current-carrying loop and understand its role in the working of electric motors and galvanometers.
Moving Coil Galvanometer: Study the construction and working of a moving coil galvanometer and learn how it can be converted into an ammeter and a voltmeter.

2.0NCERT Solutions Class 12 Physics Chapter 4 Moving Charges and Magnetism : Detailed Solutions

SOLVED EXAMPLES

A straight wire of mass 200 g and length 1.5 m carries a current of 2 A. It is suspended in mid-air by a uniform horizontal magnetic field $B$ (shown in fig.). What is the magnitude of the magnetic field?

Sol. From Lorentz formula, we find that there is an upward force F, of magnitude I/B For mid-air suspension, this must be balanced by the force due to gravity :

$mg = I / B B = \frac{mg}{I l} = \frac{0.2 \times 9.8}{2 \times 1.5} = 0.65 T$

Note that it would have been sufficient to specify $m / l$ , the mass per unit length of the wire. The earth's magnetic field is approximately $4 \times 1 0^{- 5} T$ and we have ignored it.

If the magnetic field is parallel to the positive y-axis and the charged particle is moving along the positive x-axis (Fig.), which way would the Lorentz force be for (a) an electron (negative charge), (b) a proton (positive charge).

What is the radius of the path of an electron (mass $9 \times 1 0^{- 31} kg$ and charge $1.6 \times 1 0^{- 19} C$ ) moving at a speed of $3 \times 1 0^{7} m / s$ in a magnetic field of $6 \times 1 0^{- 4} T$ perpendicular to it? What is its frequency? Calculate its energy in keV. $(1 eV = 1.6 \times 1 0^{- 19} J)$ Sol. Using equation we find, $r = mv / (qB)$

An element $Δ l = Δ x \hat{i}$ is placed at the origin and carries a large current $I = 10 A$ (figure). What is the magnetic field on the y-axis at a distance of 0.5 m. $Δ x = 1 cm$ .

Sol. $∣ d B ∣ = \frac{μ _{0}}{4 π} \frac{Id l s i n θ}{r ^{2}}$ [From Biot-Savart Law]

The direction of the field is in the +z-direction. This is so since,

$d l \times r = Δ x \hat{i} \times y \hat{j} = y Δ x (\hat{i} \times \hat{j}) = y Δ x \hat{k}$

We remind you of the following cyclic property of cross-products.

$\hat{i} \times \hat{j} = \hat{k}; \hat{j} \times \hat{k} = \hat{i}; \hat{k} \times \hat{i} = \hat{j}$

Note that the field is small in magnitude.

Sol.

$B = \frac{μ _{0} I}{4 R} = \frac{4 π \times 1 0 ^{- 7} \times 12}{4 \times 2 \times 1 0 ^{- 2}} = 6 \times 3.14 \times 1 0^{- 5} T B = 1.9 \times 1 0^{- 4} T$

Thus B is $1.9 \times 1 0^{- 4} T$ normal to the plane of the paper going into it. (c) Same magnitude of B but opposite in direction to that in (b).

Consider a tightly wound 100 turn coil of radius 10 cm. carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil ? Sol. Since the coil is tightly wound, we may take each circular element to have the same radius $R = 10 cm = 0.1 m$ . The number of turns $N = 100$ . The magnitude of the magnetic field is,

$B = \frac{μ _{0} NI}{2 R} = \frac{4 π \times 1 0 ^{- 7} \times 1 0 ^{2} \times 1}{2 \times 1 0 ^{- 1}} B = 2 π \times 1 0^{- 4} = 6.28 \times 1 0^{- 4} T$

Figure shows a long straight wire of a circular cross-section (radius a) carrying steady current I. The current I is uniformly distributed across this cross-section. Calculate the magnetic field in the region $r < a$ and $r > a$ .

Sol.

(a) Consider the case $r > a$ . The Amperian loop, labelled 2, is a circle concentric with the cross-section. For this loop.

$L = 2 π r I_{e} = Current enclosed by the loop = I$

The result is the familiar expression for a long straight wire By using Ampere's law

$B (2 π r) = μ_{0} I B_{out} = \frac{μ _{0} I}{2 π r} B_{out} \propto \frac{1}{r} (r > a)$

(b) when $r < a$ Now the current enclosed $I_{e}$ is not I, but is less than this value. Since the current distribution is uniform, the current enclosed is,

$I_{e} = I (\frac{π r _{1}^{2}}{π a ^{2}}) = \frac{I r _{1}^{2}}{a ^{2}}$

Using Ampere's law, $B (2 π r_{1}) = μ_{0} \frac{Ir _{1}^{2}}{a ^{2}}$

$B_{in} = (\frac{μ _{0} I}{2 π a ^{2}}) r_{1} B_{in} \propto (r_{1} = r < a)$

A solenoid of length 0.5 m has a radius of 1 cm and is made up of 500 turns. It carries a current of 5A. What is the magnitude of the magnetic field inside the solenoid ? Sol. The number of turns per unit length is,

$n = \frac{500}{0.5} = 1000 turns / m$

The length $l = 0.5 m$ and radius $r = 0.01 m$ . Thus, $l / a = 50$ i.e., $l ≫ a$ . Hence, we can use the long solenoid formula, i.e.

$B = μ_{0} nI B = 4 π \times 1 0^{- 7} \times 1 0^{3} \times 5 B = 6.28 \times 1 0^{- 3} T$

The horizontal component of the earth's magnetic field at a certain place is $3.0 \times 1 0^{- 5} T$ and the direction of the field is from the geographic south to the geographic north. A very long straight conductor is carrying a steady current of 1A. What is the force per unit length on it when it is placed on a horizontal table and the direction of the current is (a) east to west; (b) south to north? Sol. $F = I ℓ \times B$

$F = I / B sin θ [Magnitude of force]$

The force per unit length is

$f = F / l = IB sin θ$

(a) When the current is flowing from east to west.

$θ = 9 0^{\circ}$

Hence,

$f = IB f = 1 \times 3 \times 1 0^{- 5} = 3 \times 1 0^{- 5} Nm^{- 1}$

(b) When the current is flowing from south to north.

$θ = 0^{\circ}, f = 0$

Hence there is no force on the conductor.

A 100 turn closely wound circular coil of radius 10 cm carries a current of 3.2 A. (a) What is the field at the centre of the coil ? (b) What is the magnetic moment of this coil ? The coil is placed in a vertical plane and is free to rotate about a horizontal axis which coincides with its diameter. A uniform magnetic field of 2T in the horizontal direction exists such that initially the axis of the coil is in the direction of the field. The coil rotates through an angle of 90° under the influence of the magnetic field. (c) What are the magnitudes of the torques on the coil in the initial and final position? (d) What is the angular speed acquired by the coil when it has rotated by 90°? The moment of inertia of the coil is $0.1 kg m^{2}$ .

Sol.

(a) The magnetic field at the centre of the coil is

$B = \frac{μ _{0} NI}{2 R}$

Here,

The direction is given by the right-hand thumb rule. (b) The magnetic moment is given by below eq.,

$m = N IA = NI π r^{2} m = 100 \times 3.2 \times 3.14 \times 1 0^{- 2} = 10 A m^{2}$

The direction is once again given by the right-hand thumb rule. (c) $τ = ∣ m \times B ∣$

$τ = mB sin θ$

Initially, $θ = 0$ . Thus, initial torque $τ_{i} = 0$ . Finally, $θ = π /2$ (or 90°). Thus, final torque $τ_{f} = mB = 10 \times 2 = 20 Nm$ . (d) From Newton's second law,

$ϑ \frac{d ω}{dt} = mB sin θ$

where $ϑ$ is the moment of inertia of the coil. From chain rule,

$\frac{d ω}{dt} = \frac{d ω}{d θ} \frac{d θ}{dt} = \frac{d ω}{d θ} ω$

Using this,

$ϑ ω d ω = mB sin θ d θ$

Integrating from $θ = 0$ to $θ = π /2$ .

(a) A current-carrying circular loop lies on a smooth horizontal plane. Can a uniform magnetic field be set up in such a manner that the loop turns around itself (i.e., turns about the vertical axis). (b) A current-carrying circular loop is located in a uniform external magnetic field. If the loop is free to turn, what is its orientation of stable equilibrium? Show that in this orientation, the flux of the total field (external field + field produced by the loop) is maximum. (c) A loop of irregular shape carrying current is located in an external magnetic field. If the wire is flexible, why does it change to a circular shape?

Sol.

In the given circuit (Fig.) the current is to be measured. What is the value of the current if the ammeter shown (a) is a galvanometer with a resistance $R_{G} = 60.00Ω$ ; (b) is a galvanometer described in (a) but converted to an ammeter by a shunt resistance $r_{s} = 0.02Ω$ ; (c) is an ideal ammeter with zero resistance?

Sol.

(a) Total resistance in the circuit is, $R_{G} + 3 = 63Ω$ . Hence $I = 3/63 = 0.048 A$ (b) Resistance of the galvanometer converted to an ammeter is,

$\frac{R _{G} r _{s}}{R _{G} + r _{s}} = \frac{60Ω \times 0.02Ω}{( 60 + 0.02 ) Ω} ≃ 0.02Ω$

Total resistance in the circuit is

$0.02Ω + 3Ω = 3.02Ω. Hence, I = 3/3.02 = 0.99 A$

$I = 3/3 = 1.00 A$

EXERCISE QUESTIONS WITH SOLUTIONS

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A . What is the magnitude of the magnetic field B at the centre of the coil?

Sol. $N = 100, r = 8.0 cm = 0.08 m$ & $I = 0.4 A$ Magnitude of the magnetic field at the centre of the coil,

$B \Rightarrow B B = \frac{μ _{0} NI}{2 r} = \frac{4 π \times 1 0 ^{- 7} \times 100 \times 0.4}{2 \times 0.08} = 3.14 \times 1 0^{- 4} T$

A long straight wire carries a current of 35A. What is the magnitude of the field B at a point 20 cm from the wire?

Sol. $I = 35 A & r = 20 cm = 0.2 m$ Magnitude of the magnetic field at this point,

$B \Rightarrow B B = \frac{μ _{0} I}{2 π r} = \frac{4 π \times 1 0 ^{- 7} \times 35}{2 π \times 0.2} = 3.5 \times 1 0^{- 5} T$

A long straight wire in the horizontal plane carries a current of 50 A in north to south direction. Give the magnitude and direction of B at a point 2.5 m east of the wire. Sol. $I = 50 A, r = 2.5 m$

$B B = \frac{μ _{0} I}{2 π r} \Rightarrow B = \frac{4 π \times 1 0 ^{- 7} \times 50}{2 π \times 2.5} = 4 \times 1 0^{- 6} T$

According to the Maxwell's right hand thumb rule, the direction of the magnetic field at the given point is vertically upward.

A horizontal overhead power line carries a current of 90 A in east to west direction. What is the magnitude and direction of the magnetic field due to the current 1.5 m below the line? Sol. Current in the power line $I = 90 A$ , Point is located below the power line at distance, $r = 1.5 m$ , Hence, magnetic field at that point,

$B B = \frac{μ _{0} I}{2 π r} \Rightarrow B = \frac{4 π \times 1 0 ^{- 7} \times 90}{2 π \times 1.5} = 1.2 \times 1 0^{- 5} T$

The current is flowing from East to West. The point is below the power line. Hence, according to Maxwell's right hand thumb rule, the direction of the magnetic field is towards the South.

What is the magnitude of magnetic force per unit length on a wire carrying a current of 8A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T? Sol. $I = 8 A, B = 0.15 T, θ = 3 0^{\circ}$ . Magnetic force per unit length on the wire

$\Rightarrow \frac{F}{ℓ} = BI sin θ \frac{F}{ℓ} = 0.15 \times 8 \times sin 3 0^{\circ} = 0.6 N / m$

A 3.0 cm wire carrying a current of 10 A is placed inside a solenoid perpendicular to its axis. The magnetic field inside the solenoid is given to be 0.27 T. What is the magnetic force on the wire? Sol. $ℓ = 3 cm = 0.03 m, I = 10 A, B = 0.27 T, θ = 9 0^{\circ}$ Magnetic force exerted on the wire, $F = BI ℓ sin θ$

$F = 0.27 \times 10 \times 0.03 \times sin 9 0^{\circ} = 8.1 \times 1 0^{- 2} N$

The direction of the force can be obtained from Fleming's left hand rule.

Two long and parallel straight wires A and B carrying currents of 8.0 A and 5.0 A in the same direction are separated by a distance of 4.0 cm. Estimate the force on a 10 cm section of wire A. Sol. $I_{A} = 8.0 A, I_{B} = 5.0 A, r = 4.0 cm = 0.04 m$ Length of a section of wire A,

$L = 10 cm = 0.1 m$

Force exerted on length L due to the magnetic field, $F = \frac{μ _{0} I _{A} I _{B} L}{2 π r}$

$F = \frac{4 π \times 1 0 ^{- 7} \times 8 \times 5 \times 0.1}{2 π \times 0.04} = 2 \times 1 0^{- 5} N$

This is an attractive force normal to A towards B because the direction of the currents in the wires is the same.

A closely wound solenoid 80 cm long has 5 layers of windings of 400 turns each. The diameter of the solenoid is 1.8 cm. If the current carried is 8.0 A. estimate the magnitude of B inside the solenoid near its centre. Sol. Length of the solenoid $ℓ = 80 cm = 0.8 m$ , There are five layers of windings of 400 turns each on the solenoid. ∴ Total number of turns on the solenoid,

$N = 5 \times 400 = 2000$

Diameter of the solenoid,

$D = 1.8 cm = 0.018 m$

Current carried by the solenoid, $I = 8.0 A$ Magnitude of the magnetic field inside the solenoid near its centre is given by the relation,

$B B = \frac{μ _{0} NI}{l} = \frac{4 π \times 1 0 ^{- 7} \times 2000 \times 8}{0.8} = 2.5 \times 1 0^{- 2} T$

A square coil of side 10 cm consists of 20 turns and carries a current of 12 A. The coil is suspended vertically and the normal to the plane of the coil makes an angle of 30° with the direction of a uniform horizontal magnetic field of magnitude 0.80T. What is the magnitude of torque experienced by the coil? Sol. $ℓ = 10 cm = 0.1 m, I = 12 A, N = 20, B = 0.80 T$ Area of the square coil

$A = ℓ^{2} = 0.1 \times 0.1 = 0.01 m^{2}$

Angle made by the plane of the coil with magnetic field, $θ = 3 0^{\circ}$ Magnitude of the magnetic torque experienced by the coil in the magnetic field is given by the relation.

$τ = 0.96 N - m$

Two moving coil galvanometers, $M_{1}$ and $M_{2}$ have the following particulars:

$R_{1} = 10Ω, N_{1} = 30, A_{1} = 3.6 \times 1 0^{- 3} m^{2}, B_{1} = 0.25 T, R_{2} = 14Ω, N_{2} = 42, A_{2} = 1.8 \times 1 0^{- 3} m^{2}, B_{2} = 0.50 T$

$τ = C θ = BI NA (for M ⊥ B)$

Current sensitivity (C.S.) $= θ / I$

Voltage sensitivity (V.S.) $= θ / V = \frac{θ}{IR} = \frac{C.S.}{R}$

$\frac{( V . S . ) _{2}}{( V . S . ) _{1}} = \frac{( C.S. ) _{2}}{( C.S. ) _{1}} \times \frac{R _{1}}{R _{2}} = 1.4 \times \frac{10}{14} = 1$

In a chamber, a uniform magnetic field of 6.5 $G (1 G = 1 0^{- 4} T)$ is maintained. An electron is shot into the field with a speed of $4.8 \times 1 0^{6} ms^{- 1}$ normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit.

$(e = 1.6 \times 1 0^{- 19} C, m_{e} = 9.1 \times 1 0^{- 31} kg)$

$F_{m} F_{m} = evB = q (v \times B), (∵ v ⊥ B) = constant [Magnitude of force]$

$evB = \frac{mv ^{2}}{r} r = \frac{mv}{eB} = \frac{9.1 \times 1 0 ^{- 31} \times 4.8 \times 1 0 ^{6}}{1.6 \times 1 0 ^{- 19} \times 6.5 \times 1 0 ^{- 4}} r = 0.042 m = 4.2 cm$

In Q. 11 obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain. Sol. Frequency of revolution of electron :

$f = \frac{eB}{2 π m} = \frac{1.6 \times 1 0 ^{- 19} \times 6.5 \times 1 0 ^{- 4}}{2 \times 3.14 \times 9.1 \times 1 0 ^{- 31}} f = 18 \times 1 0^{6} rev / sec$

For uniform circular motion frequency of electron does not depends on speed of electron.

(a) A circular coil of 30 turns and radius 8.0 cm carrying a current of 6.0 A is suspended vertically in a uniform horizontal magnetic field of magnitude 1.0 T. The field lines make an angle of $6 0^{o}$ with the normal of the coil. Calculate the magnitude of the counter torque that must be applied to prevent the coil from turning. (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.) Sol. (a) Magnetic torque on coil :-

$τ τ τ = BINA sin θ = 1 \times 6 \times 30 \times 3.14 \times (8 \times 1 0^{- 2})^{2} \times sin 6 0^{\circ} = 3.1 N - m = Mechanical torque on coil = Magnetic torque on coil = 3.1 N - m$

(b) No, the answer is not changed because torque on coil is does not depends on shape of coil i.e. two different shapes have same geometrical areas.

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

Explore NCERT Solutions for Class 12 Physics chapter-wise, covering exercise answers, key formulas, and step-by-step solutions to improve concept understanding and question practice.

Chapter Number	Chapterwise NCERT Solutions
Chapter 1	Electric Charges and Fields
Chapter 2	Electrostatic Potential and Capacitance
Chapter 3	Current Electricity
Chapter 5	Magnetism and Matter
Chapter 6	Electromagnetic Induction
Chapter 7	Alternating Current
Chapter 8	Electromagnetic Waves
Chapter 9	Ray Optics and Optical Instruments
Chapter 10	Wave Optics
Chapter 11	Dual Nature of Radiation and Matter
Chapter 12	Atoms
Chapter 13	Nuclei
Chapter 14	Semiconductor Electronics: Materials, Devices and Simple Circuits

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 4

Correct Direction of Force and Field: The solutions explain the Right-Hand Thumb Rule and Fleming’s Left-Hand Rule clearly to help students find the correct direction of magnetic field and force.
Step-by-Step Derivations: Important derivations, such as the magnetic field on the axis of a circular loop and the force between two parallel current-carrying wires, are explained in simple steps.
Clear Understanding of Solenoid and Toroid: The solutions explain the magnetic field inside and outside a solenoid and the magnetic field of a toroid using Ampere’s circuital law.
Galvanometer Conversion: Step-by-step methods are provided to calculate the resistance needed to convert a galvanometer into an ammeter or voltmeter.
Use of Vector Products: The solutions explain how to use formulas such as $q (v \times B) and I (l \times B)$ to solve questions involving the direction of force and magnetic field.

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

1.0Class 12 Physics Chapter 4 Moving Charges and Magnetism: Key Concepts

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3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

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Frequently Asked Questions

Why should students choose ALLEN’s NCERT Solutions Class 12 Physics Chapter 4?

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What does NCERT Solutions Class 12 Physics Chapter 4 cover about the Biot-Savart law?

Why is Ampere’s circuital law important in NCERT Solutions Class 12 Physics Chapter 4?

What happens between two parallel current-carrying conductors according to NCERT Solutions Class 12 Physics Chapter 4?

Is NCERT Solutions Class 12 Physics Chapter 4 Moving Charges and Magnetism important for competitive exams?

Join ALLEN!

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

1.0Class 12 Physics Chapter 4 Moving Charges and Magnetism: Key Concepts

2.0NCERT Solutions Class 12 Physics Chapter 4 Moving Charges and Magnetism : Detailed Solutions

SOLVED EXAMPLES

EXERCISE QUESTIONS WITH SOLUTIONS

3.0Class 12 Physics NCERT Solutions – Chapter-wise Links

4.0Key Features of NCERT Solutions for Class 12 Physics Chapter 4

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