This chapter explores the magnetic properties of materials and their interaction with magnetic fields.
Magnetic Properties of Materials:
Diamagnetic Materials : Materials that weakly repel a magnetic field.
Paramagnetic Materials : Materials that are weakly attracted to a magnetic field.
Ferromagnetic Materials : Materials that exhibit strong magnetic properties and can be permanently magnetized.
Magnetic Hysteresis:
The lag in the magnetization of a material in response to an applied magnetic field.
It is responsible for energy loss in magnetic materials.
Electromagnets:
Devices that generate a strong magnetic field by passing an electric current through a coil of wire.
sed in various applications like motors, generators, and transformers.
Permanent Magnets:
Materials that retain their magnetic properties even after the removal of the magnetizing field.
Examples include iron, nickel, and cobalt.
This chapter delves into the fundamental concepts of magnetism, the behavior of different materials in magnetic fields, and their practical applications.
1. 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?
Ans :The torque experienced by a magnetic dipole in a uniform magnetic field is given by the equation:
τ = MB sinθ
Where:
τ is the torque
M is the magnetic moment of the dipole
B is the magnetic field strength
θ is the angle between the magnetic moment and the magnetic field direction
Rearranging the equation to solve for the magnetic moment:
M = τ / (B sinθ)
Substituting the given values:
M = (4.5 × 10⁻² J)
——————————–
(0.25 T × sin 30°)
= 0.36 J/T
Consequently, the magnitude of the magnetic moment is determined to be 0.36 Joules per Tesla
2. 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?
Ans :Given:
(M) denote Magnetic moment of the bar magnet
= 0.32 J/T
(B) denote External magnetic field
= 0.15 T
Part (a): Stable Equilibrium
The bar magnet achieves a state of stable equilibrium when its magnetic moment vector is parallel to the external magnetic field vector.. In this position, the angle (θ) between the magnetic moment vector and the external field is 0°.
(U) denote potential energy of the system is given by following formula
U = -MB cosθ
Substituting the values:
U = -(0.32 J/T)(0.15 T) cos 0° = -4.8 x 10^-2 J
Therefore, the system’s potential energy in stable equilibrium is -4.8 x 10^-2 Joules.
Part (b): Unstable Equilibrium
The bar magnet is unstable when oriented 180° to the field.
. In this position, θ = 180°.
Using the same potential energy formula:
U = -(0.32 J/T)(0.15 T) cos 180° = 4.8 x 10^-2 J
Therefore, the potential energy of the system in unstable equilibrium is 4.8 x 10^-2 Joules.
3. A closely wound solenoid of 800 turns and area of cross section 2.5 × 10-4m2 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?
Ans :A solenoid, when carrying a current, behaves like a bar magnet due to the magnetic field it generates. To find the direction of the magnetic field within a solenoid, curl your right hand’s fingers in the direction of the current flow. Your thumb will point towards the field’s direction.
(M) denote magnetic moment of a solenoid is given by following formula :
M = NIA
Where:
* N = number of turns (800)
* I = current (3.0 A)
* A = cross-sectional area (2.5 × 10⁻⁴ m²)
Substituting the values:
M = 800 × 3.0 A × 2.5 × 10⁻⁴ m² = 0.6 J/T
Therefore, the magnetic moment of the solenoid is 0.6 J/T, and it acts like a bar magnet with its north pole on one end and south pole on the other.
4. If the solenoid in Exercise 5.5 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?
Ans :The torque experienced by a solenoid in a magnetic field is given by the formula:
τ = MB sinθ
Where:
τ = torque
M = magnetic moment of the solenoid
(0.6 T⁻¹)
B = magnetic field strength
(0.25 T)
θ represents the angle formed between the axis of the solenoid and the direction of the magnetic field.
(30°)
Calculating the torque:
τ = (0.6 T⁻¹) × (0.25 T) × sin 30°
= 7.5 × 10⁻² J
consequently, the torque acting on the solenoid is 7.5 × 10⁻² Joules.
5. 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)?
Ans :(a) The work done in rotating a magnetic dipole from an initial angle θ₁ to a final angle θ₂ in a uniform magnetic field B is given by:
W = MB(cosθ₁ – cosθ₂)
(i) When the magnet is aligned with the field, θ₁ = 0°. When it’s normal to the field, θ₂ = 90°.
So, W = MB(cos 0° – cos 90°) = MB = 1.5 J/T × 0.22 T = 0.33 J
(ii) When the magnet is aligned opposite to the field, θ₂ = 180°.
W = MB(cos 0° – cos 180°) = 2MB = 2 × 1.5 J/T × 0.22 T = 0.66 J
(b) The torque on a magnetic dipole in a uniform magnetic field is given by:
τ = MB sinθ
(i) When the magnet is normal to the field, θ = 90°.
τ = MB sin 90° = MB = 1.5 J/T × 0.22 T = 0.33 Nm
(ii) When the magnet is opposite to the field, θ = 180°.
τ = MB sin 180° = 0 Nm
6. 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?
Ans :Given:
(n) denote Number of turns on the solenoid
= 2000
(A) denote Cross-sectional area of the solenoid
= 1.6 x 10⁻⁴ m²
(I) denote Current in the solenoid
= 4 A
(B) denote Magnetic field strength
= 7.5 x 10⁻² T
(θ) denote Angle between the magnetic field and the solenoid’s axis
= 30°
Part (a): Magnetic Moment
Determine magnetic moment (M) of the solenoid with following formula
M = nIA
Substituting the values:
M =
2000 × 1.6 x 10⁻⁴ m² × 4 A
= 1.28 Am²
Part (b): Torque on the Solenoid
The torque (τ) experienced by the solenoid in a magnetic field is given by:
τ = MB sinθ
Substituting the values:
τ = (1.28 Am²) × (7.5 x 10⁻² T) × sin 30° = 4.8 x 10⁻² Nm
The net force acting on the solenoid is zero owing to the uniform nature of the magnetic field. Therefore, the only force acting on the solenoid is the torque of 4.8 x 10⁻² Nm
7. 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.
Ans :(a) The magnetic field (B) at a distance (d) from the center of a bar magnet with magnetic moment (M) on its axis is given by the formula:
B = (μ₀ / 4π) * (2M / d³)
Where:
μ₀ denote permeability of free space
(4π x 10⁻⁷ Tm/A)
M represents magnetic moment of the bar magnet
(0.48 J/T)
d denote distance from the center of the magnet
(0.1 m)
Substituting the values:
B = (4π x 10⁻⁷ Tm/A) * (2 x 0.48 J/T) / (4π x (0.1 m)³)
Simplifying:
B = 0.96 x 10⁻⁴ T = 0.96 G
Therefore, the magnetic field at the given distance is 0.96 Gauss, directed along the S-N axis of the magnet.
(b)The magnetic field (B) at a distance (d) on the equatorial line of a bar magnet with magnetic moment (M) is given by:
B = (μ₀ / 4π) * (M / d³)
Where:
μ₀ denote permeability of free space
(4π x 10⁻⁷ Tm/A)
M represents magnetic moment of the bar magnet
(0.48 J/T)
d denote distance from the center of the magnet
(0.1 m)
Substituting the values:
B = (4π x 10⁻⁷ Tm/A) * (0.48 J/T) / (4π x (0.1 m)³)
Simplifying:
B = 0.48 G
Hence, the magnetic field strength at the specified distance on the equatorial line is 0.48 Gauss, aligned with the north-south axis of the magnet.
