Question 1

Q1: A short bar magnet placed with its axis at 30° with a uniform external magnetic field of 0.25 T experiences a torque of magnitude equal to 4.5 × 10⁻²...

Accepted Answer

The torque (τ) experienced by a bar magnet of magnetic moment (m) placed in a uniform external magnetic field (B) is given by the formula: τ = mB sinθ where θ is the angle between the axis of the magnet and the magnetic field. Given:    Torque, τ = 4.5 × 10⁻² J    Magnetic field, B = 0.25 T *   Angl...

Question 2

Q2: A short bar magnet of magnetic moment m = 0.32 JT⁻¹ is placed in a uniform magnetic field of 0.15 T . If the bar is free to rotate in the plane of the...

Accepted Answer

The potential energy (U) of a bar magnet with magnetic moment (m) in a uniform magnetic field (B) is given by: U = -mB cosθ where θ is the angle between the magnetic moment m and the magnetic field B. Given:    Magnetic moment, m = 0.32 J T⁻¹    Magnetic field, B = 0.15 T (a) Stable Equilibrium For...

Question 3

Q3: A closely wound solenoid of 800 turns and area of cross section 2.5 × 10⁻⁴ m² carries a current of 3.0 A. Explain the sense in which the solenoid acts...

Accepted Answer

Sense in which the solenoid acts like a bar magnet: A current-carrying solenoid behaves like a bar magnet because its magnetic field lines are similar to those of a bar magnet. The magnetic field lines emerge from one end and enter the other, forming continuous closed loops. The end from which the f...

Question 4

Q4: If the solenoid in Exercise 5.3 is free to turn about the vertical direction and a uniform horizontal magnetic field of 0.25 T is applied, what is the...

Accepted Answer

The magnitude of the torque (τ) on a magnetic dipole (like a solenoid) in a uniform magnetic field (B) is given by: τ = mB sinθ where,    m = magnetic moment of the solenoid    B = magnitude of the magnetic field    θ = angle between the axis of the solenoid and the magnetic field From the solution...

Question 5

Q5: A bar magnet of magnetic moment 1.5 J T⁻¹ lies aligned with the direction of a uniform magnetic field of 0.22 T . (a) What is the amount of work requi...

Accepted Answer

Given:    Magnetic moment, m = 1.5 J T⁻¹    Magnetic field, B = 0.22 T Initially, the magnet is aligned with the field, so the initial angle θ₁ = 0°. (a) Work Done The work done (W) in rotating the magnet from an initial angle θ₁ to a final angle θ₂ is the change in its potential energy: W = U₂ - U₁...

Question 6

Q6: A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10⁻⁴ m², carrying a current of 4.0 A , is suspended through its centre allowing...

Accepted Answer

Given:    Number of turns, N = 2000    Area of cross-section, A = 1.6 × 10⁻⁴ m²    Current, I = 4.0 A    Magnetic field, B = 7.5 × 10⁻² T    Angle, θ = 30° (a) Magnetic Moment The magnetic moment (m) of the solenoid is calculated using the formula: m = N × I × A m = 2000 × 4.0 × (1.6 × 10⁻⁴) m = 800...

Question 7

Q7: A short bar magnet has a magnetic moment of 0.48 J T⁻¹. Give the direction and magnitude of the magnetic field produced by the magnet at a distance of...

Accepted Answer

Given:    Magnetic moment, m = 0.48 J T⁻¹    Distance, r = 10 cm = 0.1 m *   We know that μ₀ / 4π = 10⁻⁷ T m A⁻¹ (a) On the axis The magnitude of the magnetic field on the axial line of a short bar magnet is given by: B_axial = (μ₀ / 4π) × (2m / r³) Substituting the values: B_axial = 10⁻⁷ × (2 × 0.4...

NCERT Solutions