Magnetism and Matter

For a bar magnet in a uniform magnetic field, the torque is

$\tau = mB\sin\theta$.

Given:
- $\tau = 4.5\times 10^{-2}\,\text{J}$
- $B = 0.25\,\text{T}$
- $\theta = 30^\circ$, so $\sin 30^\circ = \tfrac{1}{2}$

So,

$$m = \frac{\tau}{B\sin\theta} = \frac{4.5\times 10^{-2}}{0.25\times \tfrac{1}{2}}$$

$$m = \frac{4.5\times 10^{-2}}{0.125} = 0.36\,\text{J T}^{-1}$$

For a magnetic dipole in a uniform field, the potential energy is

$$U=-mB\cos\theta$$

with $m=0.32\,\text{J T}^{-1}$ and $B=0.15\,\text{T}$.

### (a) Stable equilibrium
Stable equilibrium occurs when the magnetic moment is parallel to the field, so $\theta=0^\circ$.

$$U=-mB=-0.32\times 0.15=-0.048\,\text{J}$$

### (b) Unstable equilibrium
Unstable equilibrium occurs when the magnetic moment is antiparallel to the field, so $\theta=180^\circ$.

$$U=+mB=+0.32\times 0.15=+0.048\,\text{J}$$

A closely wound current-carrying solenoid produces a magnetic field pattern similar to that of a bar magnet. Hence it acts like a magnetic dipole: one end behaves like a north pole and the other like a south pole.

Its magnetic moment is

$$m = NIA$$

where
- $N=800$,
- $I=3.0\,\text{A}$,
- $A=2.5\times 10^{-4}\,\text{m}^2$.

So,

$$m=800\times 3.0\times 2.5\times 10^{-4}$$

$$m=2400\times 2.5\times 10^{-4}=0.6\,\text{A m}^2$$

So the solenoid has the same kind of dipole field as a bar magnet, and its magnetic moment is $0.6\,\text{A m}^2$.

The torque on a magnetic dipole is

$$\tau = mB\sin\theta$$

First find the magnetic moment of the solenoid from Exercise 5.3:

$$m=NIA=800\times 3.0\times 2.5\times 10^{-4}=0.6\,\text{A m}^2$$

Now given:
- $B=0.25\,\text{T}$
- $\theta=30^\circ$, so $\sin 30^\circ=\tfrac12$

Thus,

$$\tau = 0.6\times 0.25\times \frac12$$

$$\tau = 0.075\,\text{N m}$$

Since torque has the same unit as joule per radian, its magnitude is $0.075\,\text{J}$ in the form used in the question context. The numerical magnitude is $0.075$.