Magnetism and Matter

Given:
- Angle between axis of magnet and field, $\theta = 30°$
- External magnetic field, $B = 0.25$ T
- Torque, $\tau = 4.5 \times 10^{-2}$ J

Formula used:
$$\tau = mB\sin\theta$$

Calculation:
$$m = \frac{\tau}{B\sin\theta} = \frac{4.5 \times 10^{-2}}{0.25 \times \sin 30°}$$
$$m = \frac{4.5 \times 10^{-2}}{0.25 \times 0.5} = \frac{4.5 \times 10^{-2}}{0.125}$$
$$\boxed{m = 0.36 \text{ J T}^{-1}}$$

Given:
- Magnetic moment, $m = 0.32$ J T$^{-1}$
- Magnetic field, $B = 0.15$ T

Concept: Potential energy of a magnetic dipole in a field is $U = -\mathbf{m}\cdot\mathbf{B} = -mB\cos\theta$

(a) Stable Equilibrium:
Occurs when potential energy is minimum, i.e., $\theta = 0°$ (magnetic moment aligned parallel to the field $\mathbf{B}$).
$$U_{\text{stable}} = -mB\cos 0° = -mB$$
$$U_{\text{stable}} = -(0.32)(0.15) = \boxed{-4.8 \times 10^{-2} \text{ J}}$$

(b) Unstable Equilibrium:
Occurs when potential energy is maximum, i.e., $\theta = 180°$ (magnetic moment aligned antiparallel to the field $\mathbf{B}$).
$$U_{\text{unstable}} = -mB\cos 180° = +mB$$
$$U_{\text{unstable}} = +(0.32)(0.15) = \boxed{+4.8 \times 10^{-2} \text{ J}}$$

Given:
- Number of turns, $N = 800$
- Area of cross-section, $A = 2.5 \times 10^{-4}$ m²
- Current, $I = 3.0$ A

Explanation: A current-carrying solenoid produces a magnetic field similar to that of a bar magnet. The end from which the current flows anticlockwise acts as the North pole and the end from which it flows clockwise acts as the South pole. Thus the solenoid behaves like a bar magnet with a definite north and south pole.

Formula:
$$m = NIA$$

Calculation:
$$m = 800 \times 3.0 \times 2.5 \times 10^{-4}$$
$$\boxed{m = 0.60 \text{ A m}^2 = 0.60 \text{ J T}^{-1}}$$

Given:
- Magnetic moment of solenoid (from Exercise 5.3), $m = 0.60$ J T$^{-1}$
- Magnetic field, $B = 0.25$ T
- Angle, $\theta = 30°$

Formula:
$$\tau = mB\sin\theta$$

Calculation:
$$\tau = 0.60 \times 0.25 \times \sin 30°$$
$$\tau = 0.60 \times 0.25 \times 0.5$$
$$\boxed{\tau = 7.5 \times 10^{-2} \text{ N m}}$$

Given:
- Magnetic moment, $m = 1.5$ J T$^{-1}$
- Magnetic field, $B = 0.22$ T
- Initial orientation: $\theta_1 = 0°$ (aligned with field)

Formula for work done:
$$W = U_2 - U_1 = -mB\cos\theta_2 - (-mB\cos\theta_1) = mB(\cos\theta_1 - \cos\theta_2)$$

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(a)(i) Turning to normal ($\theta_2 = 90°$):
$$W = mB(\cos 0° - \cos 90°) = mB(1 - 0) = mB$$
$$W = 1.5 \times 0.22 = \boxed{0.33 \text{ J}}$$

(a)(ii) Turning to opposite direction ($\theta_2 = 180°$):
$$W = mB(\cos 0° - \cos 180°) = mB(1 - (-1)) = 2mB$$
$$W = 2 \times 1.5 \times 0.22 = \boxed{0.66 \text{ J}}$$

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(b) Torque: $\tau = mB\sin\theta$

(i) At $\theta = 90°$:
$$\tau = mB\sin 90° = 1.5 \times 0.22 \times 1 = \boxed{0.33 \text{ N m}}$$

(ii) At $\theta = 180°$:
$$\tau = mB\sin 180° = 1.5 \times 0.22 \times 0 = \boxed{0 \text{ N m}}$$

Given:
- Number of turns, $N = 2000$
- Area of cross-section, $A = 1.6 \times 10^{-4}$ m²
- Current, $I = 4.0$ A
- Magnetic field, $B = 7.5 \times 10^{-2}$ T
- Angle between axis and field, $\theta = 30°$

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(a) Magnetic Moment:
$$m = NIA = 2000 \times 4.0 \times 1.6 \times 10^{-4}$$
$$\boxed{m = 1.28 \text{ A m}^2 = 1.28 \text{ J T}^{-1}}$$

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(b) Force and Torque:

Force: In a uniform magnetic field, the net force on a magnetic dipole is zero.
$$\boxed{F = 0}$$

Torque:
$$\tau = mB\sin\theta = 1.28 \times 7.5 \times 10^{-2} \times \sin 30°$$
$$\tau = 1.28 \times 7.5 \times 10^{-2} \times 0.5$$
$$\boxed{\tau = 4.8 \times 10^{-2} \text{ N m}}$$

Given:
- Magnetic moment, $m = 0.48$ J T$^{-1}$
- Distance, $r = 10$ cm $= 0.1$ m
- $\frac{\mu_0}{4\pi} = 10^{-7}$ T m A$^{-1}$

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(a) Field on the axial line:

Formula:
$$B_{\text{axial}} = \frac{\mu_0}{4\pi} \cdot \frac{2m}{r^3}$$

Calculation:
$$B_{\text{axial}} = 10^{-7} \times \frac{2 \times 0.48}{(0.1)^3}$$
$$B_{\text{axial}} = 10^{-7} \times \frac{0.96}{10^{-3}}$$
$$B_{\text{axial}} = 10^{-7} \times 960$$
$$\boxed{B_{\text{axial}} = 0.96 \times 10^{-4} \text{ T} = 0.96 \text{ G}}$$

Direction: Along the South-to-North direction of the magnet (i.e., along the magnetic moment $\mathbf{m}$).

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(b) Field on the equatorial line:

Formula:
$$B_{\text{equatorial}} = \frac{\mu_0}{4\pi} \cdot \frac{m}{r^3}$$

Calculation:
$$B_{\text{equatorial}} = 10^{-7} \times \frac{0.48}{(0.1)^3}$$
$$B_{\text{equatorial}} = 10^{-7} \times \frac{0.48}{10^{-3}}$$
$$B_{\text{equatorial}} = 10^{-7} \times 480$$
$$\boxed{B_{\text{equatorial}} = 0.48 \times 10^{-4} \text{ T} = 0.48 \text{ G}}$$

Direction: Opposite to the direction of the magnetic moment $\mathbf{m}$ (i.e., from North to South pole of the magnet, antiparallel to $\mathbf{m}$).