## Magnetism and Matter ncert solutions: Ncert Class 12 Physics Chapter 5 Exercise Solutions

Textbook | NCERT |

Class | Class 12 |

Subject | Physics |

Chapter | Chapter 5 |

Chapter Name | Magnetism and Matter class 12 exercise solutions |

Category | Ncert Solutions |

Medium | English |

Are you looking for Ncert Solution for Class 12 Physics Chapter 5 Magnetism and Matter? Now you can download Ncert Class 12 Physics Chapter 5 Exercise Solutions pdf from here.

### Question 5.1: A short bar magnet placed with its axis at 30º 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 5.1** To find the magnitude of the magnetic moment \((\mu\)) of the magnet, you can use the formula for the torque \((\tau\)) experienced by a magnetic dipole in a uniform magnetic field:

\(\tau = \mu B \sin \theta\)

where:

- \( \tau \) is the torque,
- \( \mu\) is the magnetic moment,
- B is the magnetic field strength,
- \( \theta\) is the angle between the magnetic moment and the magnetic field.

Given:

- \(\tau = 4.5 \times 10^{-2} \) J,
- B = 0.25 T,
- \( \theta = 30^\circ \).

First, convert the angle to radians if necessary, but since 30º = 0.5, you can use this directly.

Rearrange the formula to solve for \( \mu \):

\(\mu = \frac{\tau}{B \sin \theta} \)

Plug in the given values:

\( \mu = \frac{4.5 \times 10^{-2}}{0.25 \times \sin 30^\circ} \)

\( \mu = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5} \)

\( \mu = \frac{4.5 \times 10^{-2}}{0.125} \)

\( \mu =\) 0.36 JT^{–1}

So, the magnitude of the magnetic moment of the magnet is 0.36 JT^{–1}.

### Question 5.2: 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?

**Solution 5.2**

- Moment of the bar magnet, M = 0.32 JT
^{–1} - External magnetic field, B = 0.15 T

(a) The bar magnet is aligned along the magnetic field. This system is considered to be in stable equilibrium. Hence, the angle θ

between the bar magnet and the magnetic field is 0º

Potential energy of the system = -MB cosθ

= -0.32 **×** 0.15 cos0º

= -4.8 **×** 10^{–2} J stable.

(b) The bar magnet is oriented 180º to the magnetic field. Hence, it is in an unstable equilibrium.

θ = 180º

Potential energy = -MB cosθ

= -0.32 **×** 0.15 cos180º

= + 4.8 **×** 10^{–2} J unstable.

### Question 5.3: A closely wound solenoid of 800 turns and an area of cross-section 2.5 × 10^{–4} m^{2} 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 5.3**

- Number of turns in the solenoid, n = 800
- Area of cross-section, A = 2.5
**×**10^{–4}m^{2} - Current in the solenoid, I = 3.0 A

A current-carrying solenoid behaves like a bar magnet because a magnetic field develops along its axis, i.e., along with its length.

The magnetic moment associated with the given current-carrying solenoid is calculated as:

M = n I A

= 800 **×** 3 **×** 2.5 **×** 10^{–4}

= 0.60 JT^{–1} along the axis of the solenoid determined by the sense of flow of the current.

### Question 5.4: 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 the torque on the solenoid when its axis makes an angle of 30° with the direction of applied field?

**Solution 5.4**

- Magnetic field strength, B = 0.25 T
- Magnetic moment, M = 0.6 T
^{-1} - The angle θ between the axis of the solenoid and the direction of the applied field is 30°.
- Therefore, the torque acting on the solenoid is given as: \(\tau = MB sin \theta\)

= 0.6 **×** 0.25 sin30°

= 7.5 ×10^{–2} J

### Question 5.5: 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)?

**Solution 5.5**

**(a)** Magnetic moment, M = 1.5 J T^{–1}

Magnetic field strength, B = 0.22 T

(i) Initial angle between the axis and the magnetic field, θ_{1} = 0º

The final angle between the axis and the magnetic field, θ_{2} = 90º

The work required to make the magnetic moment normal to the direction of the magnetic field is given as:

- W = -MB(cosθ
_{2}– cosθ_{1}) - = -1.5
**×**0.22(cos 90º – cos 0º) - = -0.33 (0 – 1)
- = 0.33 J

(ii) Initial angle between the axis and the magnetic field, θ_{1} = 0º

The final angle between the axis and the magnetic field, θ_{2} = 180º

The work required to make the magnetic moment opposite to the direction of the magnetic field is given as:

- W = -MB(cosθ
_{2}– cosθ_{1}) - = -1.5
**×**0.22(cos 180º – cos 0º) - = -0.33 (-0 -1)
- = 0.66 J

**(b)** For case (i): θ = θ_{2} = 90º

∴ Torque, T = MBsinθ

- = MBsin90º
- = 1.5
**×**0.22 sin 90º - = 0.33 J

The torque tends to align the magnitude moment vector along B.

For case (ii): θ = θ_{2} = 180º

∴ Torque, T = MBsinθ

= MBsin180º = 0 J

### Question 5.6: A closely wound solenoid of 2000 turns and area of cross-section 1.6 × 10^{–4} m^{2}, 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 5.6**

- Number of turns on the solenoid, n = 2000
- Area of cross-section of the solenoid, A = 1.6 × 10
^{–4}m^{2} - Current in the solenoid, I = 4.0 A

**(a)** The magnetic moment along the axis of the solenoid is calculated as:

- M = nAI
- = 2000
**×**4**×**1.6**×**10^{–4} - = 1.28 A m
^{2}

**(b)** Magnetic field, B = 7.5 × 10^{–2} T

The angle between the magnetic field and the axis of the solenoid, θ = 30º

∴ Torque, T = MBsinθ

- = 1.28
**×**7.5**×**10^{–2}sin 30º - 0.048 Nm

Since the magnetic field is uniform, the force on the solenoid is zero. The torque on the solenoid is 0.048 Nm.

### Question 5.7: 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.

**Solution 5.7** Magnetic moment of the bar magnet, M = 0.48 J T^{–1}

**(a)** Distance, d = 10 cm = 0.1 m

The magnetic field at distance d, from the centre of the magnet on the axis, is given by the relation: \(B = \frac{\mu_{0}\;2M}{4\pi d^{3}}\)

Where,

\(\mu_{0}\) = Permeability of free space

= \(4\pi \times 10^{-7}\; TmA^{-1}\)

\(∴ B = \frac{4\pi \times 10^{-7}\times 2\times 0.48}{4\pi \times (0.1)^{3}}\)

= 0.96 **×** 10^{–4} T = 0.96 G The magnetic field is along the S-N direction.

**(b)** The magnetic field at a distance of 10 cm (i.e., d = 0.1 m) on the equatorial line of the magnet is given as:

\(B = \frac{\mu_{0}\times M}{4\pi \times d^{3}}\)

= \(\frac{4\pi \times 10^{-7}\times 0.48}{4\pi (0.1)^{3}}\)

= 0.48 G The magnetic field is along with the N–S direction.