Magnetism And MatterClass 12 Physics NCERT Solutions
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Q1EXERCISES
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 J. What is the magnitude of magnetic moment of the magnet?
Solution
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
- Angle, θ = 30°
To find the magnetic moment (m), we can rearrange the formula:
m = τ / (B sinθ)
Now, we substitute the given values into the equation:
m = (4.5 × 10⁻²) / (0.25 × sin 30°)
We know that sin 30° = 0.5.
m = (4.5 × 10⁻²) / (0.25 × 0.5)
m = (4.5 × 10⁻²) / 0.125
m = 36 × 10⁻² J T⁻¹
m = 0.36 J T⁻¹
Therefore, the magnitude of the magnetic moment of the magnet is 0.36 J T⁻¹.
Q2EXERCISES
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 field, which orientation would correspond to its (a) stable, and (b) unstable equilibrium? What is the potential energy of the magnet in each case?
Solution
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 stable equilibrium, the potential energy of the system must be minimum. This occurs when cosθ is maximum, which is +1. This happens when θ = 0°.
- Orientation: The magnet aligns itself parallel to the magnetic field, with its north pole pointing in the direction of the field.
- Potential Energy: U_stable = -mB cos(0°) U_stable = - (0.32 J T⁻¹) × (0.15 T) × 1 U_stable = -0.048 J
(b) Unstable Equilibrium
For unstable equilibrium, the potential energy of the system must be maximum. This occurs when cosθ is minimum, which is -1. This happens when θ = 180°.
- Orientation: The magnet aligns itself anti-parallel to the magnetic field, with its north pole pointing opposite to the direction of the field.
- Potential Energy: U_unstable = -mB cos(180°) U_unstable = - (0.32 J T⁻¹) × (0.15 T) × (-1) U_unstable = +0.048 J
Q3EXERCISES
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 like a bar magnet. What is its associated magnetic moment?
Solution
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 field lines emerge acts as the North pole, and the end where they enter acts as the South pole. The polarity can be determined by the right-hand thumb rule: if the fingers are curled in the direction of the current, the thumb points towards the North pole.
Calculation of Magnetic Moment:
The magnetic moment (m) of a solenoid is given by the formula:
m = N × I × A
where,
- N = total number of turns
- I = current flowing through the solenoid
- A = area of cross-section
Given:
- N = 800 turns
- A = 2.5 × 10⁻⁴ m²
- I = 3.0 A
Substituting the values into the formula:
m = 800 × 3.0 × (2.5 × 10⁻⁴)
m = 2400 × 2.5 × 10⁻⁴
m = 6000 × 10⁻⁴
m = 0.6 A m² or 0.6 J T⁻¹
Therefore, the associated magnetic moment of the solenoid is 0.6 J T⁻¹.
Q4EXERCISES
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 magnitude of torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?
Solution
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 of Exercise 5.3, we have the magnetic moment of the solenoid:
- m = 0.6 J T⁻¹
Given:
- Magnetic field, B = 0.25 T
- Angle, θ = 30°
Now, we substitute these values into the torque formula:
τ = 0.6 × 0.25 × sin 30°
We know that sin 30° = 0.5.
τ = 0.6 × 0.25 × 0.5
τ = 0.15 × 0.5
τ = 0.075 J
Therefore, the magnitude of the torque on the solenoid is 0.075 J.
Q5EXERCISES
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 required by an external torque to turn the magnet so as to align its magnetic moment: (i) normal to the field direction, (ii) opposite to the field direction?
(b)
What is the torque on the magnet in cases (i) and (ii)?
Solution
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 = (-mB cosθ₂) - (-mB cosθ₁)
W = mB (cosθ₁ - cosθ₂)
Since θ₁ = 0°, the formula becomes W = mB (cos 0° - cosθ₂) = mB (1 - cosθ₂).
(i) Normal to the field direction:
The final angle is θ₂ = 90°.
W = 1.5 × 0.22 × (1 - cos 90°)
W = 0.33 × (1 - 0)
W = 0.33 J
(ii) Opposite to the field direction:
The final angle is θ₂ = 180°.
W = 1.5 × 0.22 × (1 - cos 180°)
W = 0.33 × (1 - (-1))
W = 0.33 × 2
W = 0.66 J
(b) Torque on the Magnet
The torque (τ) on the magnet is given by the formula:
τ = mB sinθ
(i) Normal to the field direction (θ = 90°):
τ = 1.5 × 0.22 × sin 90°
τ = 0.33 × 1
τ = 0.33 J
(ii) Opposite to the field direction (θ = 180°):
τ = 1.5 × 0.22 × sin 180°
τ = 0.33 × 0
τ = 0 J
Q6EXERCISES
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 it to turn in a horizontal plane.
(a)
What is the magnetic moment associated with the solenoid?
(b)
What is the force and torque on the solenoid if a uniform horizontal magnetic field of 7.5 × 10⁻² T is set up at an angle of 30° with the axis of the solenoid?
Solution
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 = 8000 × 1.6 × 10⁻⁴
m = 12800 × 10⁻⁴
m = 1.28 A m² or 1.28 J T⁻¹
(b) Force and Torque
-
Force: Since the solenoid is placed in a uniform magnetic field, the net force on it is zero. The force on the north pole is equal and opposite to the force on the south pole, resulting in zero net force. Force = 0
-
Torque: The torque (τ) on the solenoid is given by the formula: τ = mB sinθ Substituting the values: τ = 1.28 × (7.5 × 10⁻²) × sin 30° τ = 1.28 × 7.5 × 10⁻² × 0.5 τ = 9.6 × 10⁻² × 0.5 τ = 4.8 × 10⁻² J
Therefore, the magnetic moment is 1.28 J T⁻¹, the net force is zero, and the torque is 4.8 × 10⁻² J.
Q7EXERCISES
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 10 cm from the centre of the magnet on (a) the axis, (b) the equatorial lines (normal bisector) of the magnet.
Solution
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 × 0.48) / (0.1)³
B_axial = 10⁻⁷ × 0.96 / 0.001
B_axial = 10⁻⁷ × 960
B_axial = 0.96 × 10⁻⁴ T
Direction: The direction of the magnetic field on the axial line is along the direction of the magnetic moment vector (from the South pole to the North pole of the magnet).
(b) On the equatorial lines
The magnitude of the magnetic field on the equatorial line of a short bar magnet is given by:
B_equatorial = (μ₀ / 4π) × (m / r³)
Substituting the values:
B_equatorial = 10⁻⁷ × 0.48 / (0.1)³
B_equatorial = 10⁻⁷ × 0.48 / 0.001
B_equatorial = 10⁻⁷ × 480
B_equatorial = 0.48 × 10⁻⁴ T
Direction: The direction of the magnetic field on the equatorial line is opposite to the direction of the magnetic moment vector.