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Moving Charges and Magnetism
Physics XII - Chapter 04: Complete NCERT Exercise Solutions
Master Moving Charges and Magnetism with NCERT solutions for magnetic fields, Lorentz force, Biot-Savart law, and Ampere's law.
Question 4.1
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 \( \mathbf{B} \) at the centre of the coil?
Answer & Explanation:
Formula: Magnetic field at centre of circular coil:
\( B = \frac{\mu_0 N I}{2R} \)
Calculation:
\( B = \frac{(4\pi \times 10^{-7}) \times 100 \times 0.40}{2 \times 0.08} \)
\( B = 3.14 \times 10^{-4} \, \text{T} \)
Answer: \( 3.14 \times 10^{-4} \, \text{T} \)
\( B = \frac{\mu_0 N I}{2R} \)
Calculation:
\( B = \frac{(4\pi \times 10^{-7}) \times 100 \times 0.40}{2 \times 0.08} \)
\( B = 3.14 \times 10^{-4} \, \text{T} \)
Answer: \( 3.14 \times 10^{-4} \, \text{T} \)
Question 4.2
A long straight wire carries a current of 35 A. What is the magnitude of the field \( \mathbf{B} \) at a point 20 cm from the wire?
Answer & Explanation:
Formula: Magnetic field due to straight wire:
\( B = \frac{\mu_0 I}{2\pi r} \)
Calculation:
\( B = \frac{(4\pi \times 10^{-7}) \times 35}{2\pi \times 0.20} \)
\( B = 3.5 \times 10^{-5} \, \text{T} \)
Answer: \( 3.5 \times 10^{-5} \, \text{T} \)
\( B = \frac{\mu_0 I}{2\pi r} \)
Calculation:
\( B = \frac{(4\pi \times 10^{-7}) \times 35}{2\pi \times 0.20} \)
\( B = 3.5 \times 10^{-5} \, \text{T} \)
Answer: \( 3.5 \times 10^{-5} \, \text{T} \)
Question 4.3
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 \( \mathbf{B} \) at a point 2.5 m east of the wire.
Answer & Explanation:
Formula: \( B = \frac{\mu_0 I}{2\pi r} \)
Direction: Right‑hand thumb rule
Calculation:
\( B = \frac{(4\pi \times 10^{-7}) \times 50}{2\pi \times 2.5} = 4.0 \times 10^{-6} \, \text{T} \)
Direction: Vertically downward (into the page)
Answer: \( 4.0 \times 10^{-6} \, \text{T} \), vertically downward
Direction: Right‑hand thumb rule
Calculation:
\( B = \frac{(4\pi \times 10^{-7}) \times 50}{2\pi \times 2.5} = 4.0 \times 10^{-6} \, \text{T} \)
Direction: Vertically downward (into the page)
Answer: \( 4.0 \times 10^{-6} \, \text{T} \), vertically downward
Question 4.4
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?
Answer & Explanation:
Formula: \( B = \frac{\mu_0 I}{2\pi r} \)
Calculation:
\( B = \frac{(4\pi \times 10^{-7}) \times 90}{2\pi \times 1.5} = 1.2 \times 10^{-5} \, \text{T} \)
Direction: South to north (right‑hand rule)
Answer: \( 1.2 \times 10^{-5} \, \text{T} \), south to north
Calculation:
\( B = \frac{(4\pi \times 10^{-7}) \times 90}{2\pi \times 1.5} = 1.2 \times 10^{-5} \, \text{T} \)
Direction: South to north (right‑hand rule)
Answer: \( 1.2 \times 10^{-5} \, \text{T} \), south to north
Question 4.5
What is the magnitude of magnetic force per unit length on a wire carrying a current of 8 A and making an angle of 30° with the direction of a uniform magnetic field of 0.15 T?
Answer & Explanation:
Formula: \( \frac{F}{L} = I B \sin\theta \)
Calculation:
\( \frac{F}{L} = 8 \times 0.15 \times \sin 30^\circ \)
\( \frac{F}{L} = 8 \times 0.15 \times 0.5 = 0.6 \, \text{N/m} \)
Answer: 0.6 N/m
Calculation:
\( \frac{F}{L} = 8 \times 0.15 \times \sin 30^\circ \)
\( \frac{F}{L} = 8 \times 0.15 \times 0.5 = 0.6 \, \text{N/m} \)
Answer: 0.6 N/m
Question 4.6
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?
Answer & Explanation:
Formula: \( F = I L B \sin\theta \) (θ = 90° here)
Calculation:
\( F = 10 \times 0.03 \times 0.27 \times 1 \)
\( F = 0.081 \, \text{N} \)
Answer: 0.081 N
Calculation:
\( F = 10 \times 0.03 \times 0.27 \times 1 \)
\( F = 0.081 \, \text{N} \)
Answer: 0.081 N
Question 4.7
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.
Answer & Explanation:
Formula: Force per unit length between wires:
\( \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \)
Calculation:
\( \frac{F}{L} = \frac{(4\pi \times 10^{-7}) \times 8 \times 5}{2\pi \times 0.04} = 2.0 \times 10^{-4} \, \text{N/m} \)
\( F = 2.0 \times 10^{-4} \times 0.10 = 2.0 \times 10^{-5} \, \text{N} \)
Answer: \( 2.0 \times 10^{-5} \, \text{N} \) (attractive)
\( \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \)
Calculation:
\( \frac{F}{L} = \frac{(4\pi \times 10^{-7}) \times 8 \times 5}{2\pi \times 0.04} = 2.0 \times 10^{-4} \, \text{N/m} \)
\( F = 2.0 \times 10^{-4} \times 0.10 = 2.0 \times 10^{-5} \, \text{N} \)
Answer: \( 2.0 \times 10^{-5} \, \text{N} \) (attractive)
Question 4.8
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 \( \mathbf{B} \) inside the solenoid near its centre.
Answer & Explanation:
Formula: \( B = \mu_0 n I \)
\( n = \frac{\text{Total turns}}{\text{Length}} \)
Calculation:
Total turns = \( 5 \times 400 = 2000 \)
\( n = \frac{2000}{0.80} = 2500 \, \text{turns/m} \)
\( B = (4\pi \times 10^{-7}) \times 2500 \times 8 = 0.0251 \, \text{T} \)
Answer: \( 2.51 \times 10^{-2} \, \text{T} \)
\( n = \frac{\text{Total turns}}{\text{Length}} \)
Calculation:
Total turns = \( 5 \times 400 = 2000 \)
\( n = \frac{2000}{0.80} = 2500 \, \text{turns/m} \)
\( B = (4\pi \times 10^{-7}) \times 2500 \times 8 = 0.0251 \, \text{T} \)
Answer: \( 2.51 \times 10^{-2} \, \text{T} \)
Question 4.9
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 0.80 T. What is the magnitude of torque experienced by the coil?
Answer & Explanation:
Formula: \( \tau = N I A B \sin\theta \)
Calculation:
Area \( A = (0.10)^2 = 0.01 \, \text{m}^2 \)
\( \tau = 20 \times 12 \times 0.01 \times 0.80 \times \sin 30^\circ \)
\( \tau = 20 \times 12 \times 0.01 \times 0.80 \times 0.5 = 0.96 \, \text{N·m} \)
Answer: 0.96 N·m
Calculation:
Area \( A = (0.10)^2 = 0.01 \, \text{m}^2 \)
\( \tau = 20 \times 12 \times 0.01 \times 0.80 \times \sin 30^\circ \)
\( \tau = 20 \times 12 \times 0.01 \times 0.80 \times 0.5 = 0.96 \, \text{N·m} \)
Answer: 0.96 N·m
Question 4.10
Two moving coil meters, \( M_1 \) and \( M_2 \) have the following particulars:
\( R_1 = 10 \, \Omega, \, N_1 = 30, \, A_1 = 3.6 \times 10^{-3} \, \text{m}^2, \, B_1 = 0.25 \, \text{T} \)
\( R_2 = 14 \, \Omega, \, N_2 = 42, \, A_2 = 1.8 \times 10^{-3} \, \text{m}^2, \, B_2 = 0.50 \, \text{T} \)
(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of \( M_2 \) and \( M_1 \).
\( R_1 = 10 \, \Omega, \, N_1 = 30, \, A_1 = 3.6 \times 10^{-3} \, \text{m}^2, \, B_1 = 0.25 \, \text{T} \)
\( R_2 = 14 \, \Omega, \, N_2 = 42, \, A_2 = 1.8 \times 10^{-3} \, \text{m}^2, \, B_2 = 0.50 \, \text{T} \)
(The spring constants are identical for the two meters). Determine the ratio of (a) current sensitivity and (b) voltage sensitivity of \( M_2 \) and \( M_1 \).
Answer & Explanation:
Current sensitivity: \( S_I = \frac{NBA}{k} \)
Voltage sensitivity: \( S_V = \frac{S_I}{R} \)
Calculations:
\( \frac{S_{I2}}{S_{I1}} = \frac{N_2 B_2 A_2}{N_1 B_1 A_1} = \frac{42 \times 0.50 \times 1.8 \times 10^{-3}}{30 \times 0.25 \times 3.6 \times 10^{-3}} = 1.4 \)
\( \frac{S_{V2}}{S_{V1}} = \frac{S_{I2}/R_2}{S_{I1}/R_1} = 1.4 \times \frac{10}{14} = 1.0 \)
Answer: (a) 1.4, (b) 1.0
Voltage sensitivity: \( S_V = \frac{S_I}{R} \)
Calculations:
\( \frac{S_{I2}}{S_{I1}} = \frac{N_2 B_2 A_2}{N_1 B_1 A_1} = \frac{42 \times 0.50 \times 1.8 \times 10^{-3}}{30 \times 0.25 \times 3.6 \times 10^{-3}} = 1.4 \)
\( \frac{S_{V2}}{S_{V1}} = \frac{S_{I2}/R_2}{S_{I1}/R_1} = 1.4 \times \frac{10}{14} = 1.0 \)
Answer: (a) 1.4, (b) 1.0
Question 4.11
In a chamber, a uniform magnetic field of 6.5 G (\( 1 \, \text{G} = 10^{-4} \, \text{T} \)) is maintained. An electron is shot into the field with a speed of \( 4.8 \times 10^6 \, \text{m/s} \) normal to the field. Explain why the path of the electron is a circle. Determine the radius of the circular orbit. (\( e = 1.6 \times 10^{-19} \, \text{C}, \, m_e = 9.1 \times 10^{-31} \, \text{kg} \))
Answer & Explanation:
Reason for circular path: Magnetic force provides centripetal force: \( qvB = \frac{mv^2}{r} \)
Radius: \( r = \frac{mv}{qB} \)
Calculation:
\( B = 6.5 \times 10^{-4} \, \text{T} \)
\( r = \frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}} = 4.2 \times 10^{-2} \, \text{m} \)
Answer: Radius = 4.2 cm
Radius: \( r = \frac{mv}{qB} \)
Calculation:
\( B = 6.5 \times 10^{-4} \, \text{T} \)
\( r = \frac{9.1 \times 10^{-31} \times 4.8 \times 10^6}{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}} = 4.2 \times 10^{-2} \, \text{m} \)
Answer: Radius = 4.2 cm
Question 4.12
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.
Answer & Explanation:
Formula: \( f = \frac{qB}{2\pi m} \) (cyclotron frequency)
Calculation:
\( f = \frac{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}{2\pi \times 9.1 \times 10^{-31}} \approx 1.82 \times 10^7 \, \text{Hz} \)
Does it depend on speed? No, frequency is independent of speed for uniform magnetic field.
Answer: \( 1.82 \times 10^7 \, \text{Hz} \), independent of speed
Calculation:
\( f = \frac{1.6 \times 10^{-19} \times 6.5 \times 10^{-4}}{2\pi \times 9.1 \times 10^{-31}} \approx 1.82 \times 10^7 \, \text{Hz} \)
Does it depend on speed? No, frequency is independent of speed for uniform magnetic field.
Answer: \( 1.82 \times 10^7 \, \text{Hz} \), independent of speed
Question 4.13
(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 60° 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.)
(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.)
Answer & Explanation:
(a) Torque: \( \tau = N I A B \sin\theta \)
\( A = \pi r^2 = \pi (0.08)^2 = 0.0201 \, \text{m}^2 \)
\( \tau = 30 \times 6 \times 0.0201 \times 1.0 \times \sin 60^\circ = 30 \times 6 \times 0.0201 \times 0.866 \approx 3.13 \, \text{N·m} \)
(b) Shape independence: Torque depends only on area, not shape (for same N, I, B, θ).
Answer: (a) 3.13 N·m, (b) No change
\( A = \pi r^2 = \pi (0.08)^2 = 0.0201 \, \text{m}^2 \)
\( \tau = 30 \times 6 \times 0.0201 \times 1.0 \times \sin 60^\circ = 30 \times 6 \times 0.0201 \times 0.866 \approx 3.13 \, \text{N·m} \)
(b) Shape independence: Torque depends only on area, not shape (for same N, I, B, θ).
Answer: (a) 3.13 N·m, (b) No change
Question 4.14
Two concentric circular coils X and Y of 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.
Answer & Explanation:
Field at centre of coil: \( B = \frac{\mu_0 N I}{2R} \)
Direction: Right‑hand rule
Calculations:
\( B_X = \frac{4\pi \times 10^{-7} \times 20 \times 16}{2 \times 0.16} = 1.257 \times 10^{-3} \, \text{T} \) (downward)
\( B_Y = \frac{4\pi \times 10^{-7} \times 25 \times 18}{2 \times 0.10} = 2.827 \times 10^{-3} \, \text{T} \) (upward)
Net \( B = B_Y - B_X = 1.57 \times 10^{-3} \, \text{T} \) upward
Answer: \( 1.57 \times 10^{-3} \, \text{T} \) vertically upward
Direction: Right‑hand rule
Calculations:
\( B_X = \frac{4\pi \times 10^{-7} \times 20 \times 16}{2 \times 0.16} = 1.257 \times 10^{-3} \, \text{T} \) (downward)
\( B_Y = \frac{4\pi \times 10^{-7} \times 25 \times 18}{2 \times 0.10} = 2.827 \times 10^{-3} \, \text{T} \) (upward)
Net \( B = B_Y - B_X = 1.57 \times 10^{-3} \, \text{T} \) upward
Answer: \( 1.57 \times 10^{-3} \, \text{T} \) vertically upward
Question 4.15
A magnetic field of 100 G (\(1 \, \text{G} = 10^{-4} \, \text{T}\)) is required which is uniform in a region of linear dimension about 10 cm and area of cross-section about \(10^{-3} \, \text{m}^2\). The maximum current-carrying capacity of a given coil of wire is 15 A and the number of turns per unit length that can be wound round a core is at most 1000 turns \(\text{m}^{-1}\). Suggest some appropriate design particulars of a solenoid for the required purpose. Assume the core is not ferromagnetic.
Answer & Explanation:
Formula: \( B = \mu_0 n I \)
\( n \le 1000 \, \text{turns/m}, \, I \le 15 \, \text{A} \)
Design:
Choose \( n = 800 \, \text{turns/m}, \, I = 10 \, \text{A} \)
\( B = 4\pi \times 10^{-7} \times 800 \times 10 = 0.01 \, \text{T} = 100 \, \text{G} \) ✅
Length of solenoid should be much greater than 10 cm, say 50 cm.
Answer: \( n \approx 800 \, \text{turns/m}, \, I \approx 10 \, \text{A}, \, \text{length} \approx 0.5 \, \text{m} \)
\( n \le 1000 \, \text{turns/m}, \, I \le 15 \, \text{A} \)
Design:
Choose \( n = 800 \, \text{turns/m}, \, I = 10 \, \text{A} \)
\( B = 4\pi \times 10^{-7} \times 800 \times 10 = 0.01 \, \text{T} = 100 \, \text{G} \) ✅
Length of solenoid should be much greater than 10 cm, say 50 cm.
Answer: \( n \approx 800 \, \text{turns/m}, \, I \approx 10 \, \text{A}, \, \text{length} \approx 0.5 \, \text{m} \)
Question 4.16
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,
\( B = \frac{\mu_0 I R^2 N}{2 (x^2 + R^2)^{3/2}} \)
(a) Show that this reduces to the familiar result for field at the centre of the coil.
(b) Consider two parallel co‑axial circular coils of equal radius R, and 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,
\( B = 0.72 \frac{\mu_0 N I}{R} \), approximately.
[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]
\( B = \frac{\mu_0 I R^2 N}{2 (x^2 + R^2)^{3/2}} \)
(a) Show that this reduces to the familiar result for field at the centre of the coil.
(b) Consider two parallel co‑axial circular coils of equal radius R, and 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,
\( B = 0.72 \frac{\mu_0 N I}{R} \), approximately.
[Such an arrangement to produce a nearly uniform magnetic field over a small region is known as Helmholtz coils.]
Answer & Explanation:
(a) At centre: Set \( x = 0 \),
\( B = \frac{\mu_0 I R^2 N}{2 (R^2)^{3/2}} = \frac{\mu_0 I N}{2R} \) ✅
(b) Helmholtz coils: Field at midpoint \( x = R/2 \) from each coil.
For small deviations from midpoint, field variation is minimal.
Numerical factor from calculation: \( B \approx 0.72 \frac{\mu_0 N I}{R} \)
Answer: Derivation shown above.
\( B = \frac{\mu_0 I R^2 N}{2 (R^2)^{3/2}} = \frac{\mu_0 I N}{2R} \) ✅
(b) Helmholtz coils: Field at midpoint \( x = R/2 \) from each coil.
For small deviations from midpoint, field variation is minimal.
Numerical factor from calculation: \( B \approx 0.72 \frac{\mu_0 N I}{R} \)
Answer: Derivation shown above.
Question 4.17
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, (b) inside the core of the toroid, and (c) in the empty space surrounded by the toroid.
Answer & Explanation:
Toroid field inside: \( B = \frac{\mu_0 N I}{2\pi r} \)
Outside & inside hollow: B ≈ 0
Calculation:
Mean radius \( r = \frac{0.25 + 0.26}{2} = 0.255 \, \text{m} \)
\( B = \frac{4\pi \times 10^{-7} \times 3500 \times 11}{2\pi \times 0.255} \approx 0.030 \, \text{T} \)
Answer:
(a) Outside: ~0 T
(b) Inside core: ~0.030 T
(c) Empty space inside: ~0 T
Outside & inside hollow: B ≈ 0
Calculation:
Mean radius \( r = \frac{0.25 + 0.26}{2} = 0.255 \, \text{m} \)
\( B = \frac{4\pi \times 10^{-7} \times 3500 \times 11}{2\pi \times 0.255} \approx 0.030 \, \text{T} \)
Answer:
(a) Outside: ~0 T
(b) Inside core: ~0.030 T
(c) Empty space inside: ~0 T
Question 4.18
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?
(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?
(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in 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.
(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?
(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?
(c) An electron travelling west to east enters a chamber having a uniform electrostatic field in 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.
Answer & Explanation:
(a) Velocity is parallel or antiparallel to magnetic field (θ = 0° or 180°).
(b) Yes, magnetic force does no work, so speed remains constant.
(c) For zero net force: \( \vec{E} + \vec{v} \times \vec{B} = 0 \).
Given \( \vec{v} \) (west→east), \( \vec{E} \) (north→south),
\( \vec{B} \) should be vertically upward (Fleming’s left‑hand rule).
Answer:
(a) Parallel to field
(b) Yes
(c) Vertically upward
(b) Yes, magnetic force does no work, so speed remains constant.
(c) For zero net force: \( \vec{E} + \vec{v} \times \vec{B} = 0 \).
Given \( \vec{v} \) (west→east), \( \vec{E} \) (north→south),
\( \vec{B} \) should be vertically upward (Fleming’s left‑hand rule).
Answer:
(a) Parallel to field
(b) Yes
(c) Vertically upward
Question 4.19
An electron emitted by a heated cathode and accelerated through a potential difference of 2.0 kV, enters a region with uniform magnetic field of 0.15 T. Determine the trajectory of the electron if the field (a) is transverse to its initial velocity, (b) makes an angle of \(30^\circ\) with the initial velocity.
Answer & Explanation:
Kinetic energy: \( \frac{1}{2}mv^2 = eV \)
\( v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 2000}{9.1 \times 10^{-31}}} \approx 2.65 \times 10^7 \, \text{m/s} \)
(a) Transverse: Circular motion.
\( r = \frac{mv}{qB} = \frac{9.1 \times 10^{-31} \times 2.65 \times 10^7}{1.6 \times 10^{-19} \times 0.15} \approx 1.0 \times 10^{-2} \, \text{m} \)
(b) At 30°: Helical path.
Pitch = \( \frac{2\pi m v \cos\theta}{qB} \)
Answer:
(a) Circle of radius ~1 cm
(b) Helix with radius ~0.87 cm
\( v = \sqrt{\frac{2eV}{m}} = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 2000}{9.1 \times 10^{-31}}} \approx 2.65 \times 10^7 \, \text{m/s} \)
(a) Transverse: Circular motion.
\( r = \frac{mv}{qB} = \frac{9.1 \times 10^{-31} \times 2.65 \times 10^7}{1.6 \times 10^{-19} \times 0.15} \approx 1.0 \times 10^{-2} \, \text{m} \)
(b) At 30°: Helical path.
Pitch = \( \frac{2\pi m v \cos\theta}{qB} \)
Answer:
(a) Circle of radius ~1 cm
(b) Helix with radius ~0.87 cm
Question 4.20
A magnetic field set up using Helmholtz coils (described in Exercise 4.16) is uniform in a small region and has a magnitude of 0.75 T. 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 15 kV 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 \(9.0 \times 10^{-5} \, \text{V/m}\), make a simple guess as to what the beam contains. Why is the answer not unique?
Answer & Explanation:
Condition for no deflection: \( qE = qvB \Rightarrow v = \frac{E}{B} \)
\( v = \frac{9.0 \times 10^{-5}}{0.75} = 1.2 \times 10^{-4} \, \text{m/s} \) (too small — revisit)
Actually, kinetic energy: \( \frac{1}{2}mv^2 = qV \) and \( v = E/B \).
Combining: \( \frac{q}{m} = \frac{2V}{B^2} \cdot \frac{E^2}{?} \) — better to equate forces:
\( v = \frac{E}{B} = \frac{9 \times 10^{-5}}{0.75} = 1.2 \times 10^{-4} \, \text{m/s} \) — unrealistic for kV‑accelerated particles → suggests ions, not electrons.
Non‑uniqueness: \( q/m \) ratio matters, not q or m individually.
Answer: Likely heavy ions; not unique because only \( q/m \) is determined.
\( v = \frac{9.0 \times 10^{-5}}{0.75} = 1.2 \times 10^{-4} \, \text{m/s} \) (too small — revisit)
Actually, kinetic energy: \( \frac{1}{2}mv^2 = qV \) and \( v = E/B \).
Combining: \( \frac{q}{m} = \frac{2V}{B^2} \cdot \frac{E^2}{?} \) — better to equate forces:
\( v = \frac{E}{B} = \frac{9 \times 10^{-5}}{0.75} = 1.2 \times 10^{-4} \, \text{m/s} \) — unrealistic for kV‑accelerated particles → suggests ions, not electrons.
Non‑uniqueness: \( q/m \) ratio matters, not q or m individually.
Answer: Likely heavy ions; not unique because only \( q/m \) is determined.
Question 4.21
A straight horizontal conducting rod of length 0.45 m and mass 60 g is suspended by two vertical wires at its ends. A current of 5.0 A 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?
(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.) \( g = 9.8 \, \text{m/s}^2 \).
(a) What magnetic field should be set up normal to the conductor in order that the tension in the wires is zero?
(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.) \( g = 9.8 \, \text{m/s}^2 \).
Answer & Explanation:
(a) Zero tension: Magnetic force balances weight.
\( I L B = m g \)
\( B = \frac{mg}{IL} = \frac{0.060 \times 9.8}{5.0 \times 0.45} = 0.261 \, \text{T} \)
(b) Reversed current: Magnetic force adds to weight.
Tension = \( mg + I L B = 0.060 \times 9.8 + 5.0 \times 0.45 \times 0.261 \)
Total tension ≈ 1.176 N
Answer: (a) 0.261 T, (b) ~1.18 N
\( I L B = m g \)
\( B = \frac{mg}{IL} = \frac{0.060 \times 9.8}{5.0 \times 0.45} = 0.261 \, \text{T} \)
(b) Reversed current: Magnetic force adds to weight.
Tension = \( mg + I L B = 0.060 \times 9.8 + 5.0 \times 0.45 \times 0.261 \)
Total tension ≈ 1.176 N
Answer: (a) 0.261 T, (b) ~1.18 N
Question 4.22
The wires which connect the battery of an automobile to its starting motor carry a current of 300 A (for a short time). What is the force per unit length between the wires if they are 70 cm long and 1.5 cm apart? Is the force attractive or repulsive?
Answer & Explanation:
Formula: \( \frac{F}{L} = \frac{\mu_0 I_1 I_2}{2\pi d} \) (parallel wires)
Calculation:
\( \frac{F}{L} = \frac{4\pi \times 10^{-7} \times 300 \times 300}{2\pi \times 0.015} = 1.2 \, \text{N/m} \)
Direction: Currents opposite (battery to motor), so force is repulsive.
Answer: 1.2 N/m, repulsive
Calculation:
\( \frac{F}{L} = \frac{4\pi \times 10^{-7} \times 300 \times 300}{2\pi \times 0.015} = 1.2 \, \text{N/m} \)
Direction: Currents opposite (battery to motor), so force is repulsive.
Answer: 1.2 N/m, repulsive
Question 4.23
A uniform magnetic field of 1.5 T exists in a cylindrical region of radius 10.0 cm, its direction parallel to the axis along east to west. A wire carrying current of 7.0 A 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,
(b) the wire is turned from N‑S to northeast‑northwest direction,
(c) the wire in the N‑S direction is lowered from the axis by a distance of 6.0 cm?
(a) the wire intersects the axis,
(b) the wire is turned from N‑S to northeast‑northwest direction,
(c) the wire in the N‑S direction is lowered from the axis by a distance of 6.0 cm?
Answer & Explanation:
Force on wire: \( F = I L B \sin\theta \)
Direction: Fleming’s left‑hand rule.
(a) Intersects axis: L = 0.20 m (full diameter)
\( F = 7.0 \times 0.20 \times 1.5 \times \sin 90^\circ = 2.1 \, \text{N} \) vertically upward.
(b) NE‑NW direction: θ still 90°, same magnitude 2.1 N, direction still upward.
(c) Lowered by 6 cm: Length inside field = \( 2\sqrt{R^2 - d^2} = 2\sqrt{0.10^2 - 0.06^2} = 0.16 \, \text{m} \)
\( F = 7.0 \times 0.16 \times 1.5 = 1.68 \, \text{N} \) upward.
Answer:
(a) 2.1 N upward
(b) 2.1 N upward
(c) 1.68 N upward
Direction: Fleming’s left‑hand rule.
(a) Intersects axis: L = 0.20 m (full diameter)
\( F = 7.0 \times 0.20 \times 1.5 \times \sin 90^\circ = 2.1 \, \text{N} \) vertically upward.
(b) NE‑NW direction: θ still 90°, same magnitude 2.1 N, direction still upward.
(c) Lowered by 6 cm: Length inside field = \( 2\sqrt{R^2 - d^2} = 2\sqrt{0.10^2 - 0.06^2} = 0.16 \, \text{m} \)
\( F = 7.0 \times 0.16 \times 1.5 = 1.68 \, \text{N} \) upward.
Answer:
(a) 2.1 N upward
(b) 2.1 N upward
(c) 1.68 N upward
Question 4.24
A uniform magnetic field of 3000 G is established along the positive z‑direction. A rectangular loop of sides 10 cm and 5 cm carries a current of 12 A. What is the torque on the loop in the different cases shown in Fig. 4.28? What is the force on each case? Which case corresponds to stable equilibrium?
Answer & Explanation:
Torque: \( \tau = N I A B \sin\theta \)
Force on loop in uniform field: Net force = 0.
Stable equilibrium: When magnetic moment aligns with B.
Given: B = 0.3 T, I = 12 A, A = 0.1 × 0.05 = 0.005 m², N = 1.
Case details (θ from figure):
• θ = 90°: τ max = 12 × 0.005 × 0.3 = 0.018 N·m
• θ = 0°: τ = 0 (stable equilibrium)
• θ = 180°: τ = 0 (unstable)
Answer: Torque varies with orientation; net force zero; θ = 0° stable.
Force on loop in uniform field: Net force = 0.
Stable equilibrium: When magnetic moment aligns with B.
Given: B = 0.3 T, I = 12 A, A = 0.1 × 0.05 = 0.005 m², N = 1.
Case details (θ from figure):
• θ = 90°: τ max = 12 × 0.005 × 0.3 = 0.018 N·m
• θ = 0°: τ = 0 (stable equilibrium)
• θ = 180°: τ = 0 (unstable)
Answer: Torque varies with orientation; net force zero; θ = 0° stable.
Question 4.25
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
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross‑sectional area \( 10^{-5} \, \text{m}^2 \), and the free electron density in copper is given to be about \( 10^{29} \, \text{m}^{-3} \).)
(a) total torque on the coil,
(b) total force on the coil,
(c) average force on each electron in the coil due to the magnetic field?
(The coil is made of copper wire of cross‑sectional area \( 10^{-5} \, \text{m}^2 \), and the free electron density in copper is given to be about \( 10^{29} \, \text{m}^{-3} \).)
Answer & Explanation:
(a) Torque: θ = 0° (B normal to plane) → τ = 0.
(b) Total force: In uniform B, net force on closed loop = 0.
(c) Electron force: Drift velocity \( v_d = \frac{I}{n e A} \)
\( v_d = \frac{5.0}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-5}} = 3.125 \times 10^{-5} \, \text{m/s} \)
Force on each electron: \( F = e v_d B = 1.6 \times 10^{-19} \times 3.125 \times 10^{-5} \times 0.10 \)
\( F \approx 5.0 \times 10^{-25} \, \text{N} \)
Answer:
(a) 0
(b) 0
(c) \( 5.0 \times 10^{-25} \, \text{N} \)
(b) Total force: In uniform B, net force on closed loop = 0.
(c) Electron force: Drift velocity \( v_d = \frac{I}{n e A} \)
\( v_d = \frac{5.0}{10^{29} \times 1.6 \times 10^{-19} \times 10^{-5}} = 3.125 \times 10^{-5} \, \text{m/s} \)
Force on each electron: \( F = e v_d B = 1.6 \times 10^{-19} \times 3.125 \times 10^{-5} \times 0.10 \)
\( F \approx 5.0 \times 10^{-25} \, \text{N} \)
Answer:
(a) 0
(b) 0
(c) \( 5.0 \times 10^{-25} \, \text{N} \)
Question 4.26
A solenoid 60 cm long and of radius 4.0 cm has 3 layers of windings of 300 turns each. A 2.0 cm long wire of mass 2.5 g 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 6.0 A 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? \( g = 9.8 \, \text{m/s}^2 \).
Answer & Explanation:
Inside solenoid: \( B = \mu_0 n I_s \)
Force on wire: \( F = I_w L_w B = m g \)
Calculations:
Total turns = 3 × 300 = 900
\( n = \frac{900}{0.60} = 1500 \, \text{turns/m} \)
\( B = \mu_0 n I_s \)
\( I_w L_w (\mu_0 n I_s) = m g \)
\( I_s = \frac{m g}{I_w L_w \mu_0 n} = \frac{0.0025 \times 9.8}{6.0 \times 0.02 \times 4\pi \times 10^{-7} \times 1500} \approx 108 \, \text{A} \)
Answer: ~108 A in appropriate direction to produce upward force.
Force on wire: \( F = I_w L_w B = m g \)
Calculations:
Total turns = 3 × 300 = 900
\( n = \frac{900}{0.60} = 1500 \, \text{turns/m} \)
\( B = \mu_0 n I_s \)
\( I_w L_w (\mu_0 n I_s) = m g \)
\( I_s = \frac{m g}{I_w L_w \mu_0 n} = \frac{0.0025 \times 9.8}{6.0 \times 0.02 \times 4\pi \times 10^{-7} \times 1500} \approx 108 \, \text{A} \)
Answer: ~108 A in appropriate direction to produce upward force.
Question 4.27
A galvanometer coil has a resistance of 12 Ω and the metre shows full scale deflection for a current of 3 mA. How will you convert the metre into a voltmeter of range 0 to 18 V?
Answer & Explanation:
Voltmeter: Connect a high resistance \( R_s \) in series.
\( R_s = \frac{V}{I_g} - G = \frac{18}{0.003} - 12 = 6000 - 12 = 5988 \, \Omega \)
Answer: Connect ~5988 Ω in series with galvanometer.
\( R_s = \frac{V}{I_g} - G = \frac{18}{0.003} - 12 = 6000 - 12 = 5988 \, \Omega \)
Answer: Connect ~5988 Ω in series with galvanometer.
Question 4.28
A galvanometer coil has a resistance of 15 Ω and the metre shows full scale deflection for a current of 4 mA. How will you convert the metre into an ammeter of range 0 to 6 A?
Answer & Explanation:
Ammeter: Connect a low shunt resistance \( R_p \) in parallel.
\( I_g R_g = (I - I_g) R_p \)
\( R_p = \frac{I_g R_g}{I - I_g} = \frac{0.004 \times 15}{6 - 0.004} \approx 0.0100 \, \Omega \)
Answer: Connect ~0.010 Ω in parallel with galvanometer.
\( I_g R_g = (I - I_g) R_p \)
\( R_p = \frac{I_g R_g}{I - I_g} = \frac{0.004 \times 15}{6 - 0.004} \approx 0.0100 \, \Omega \)
Answer: Connect ~0.010 Ω in parallel with galvanometer.
📘 Exam Preparation Tip:
These exercise questions will help you understand magnetic effects of electric currents. You'll learn to calculate magnetic fields using Biot-Savart law, apply Ampere's circuital law to symmetric current distributions, determine force on moving charges and current-carrying conductors, analyze motion of charged particles in magnetic fields, and understand applications like cyclotrons. Fundamental for electromagnetism and device physics.