Q.A bar magnet is placed in a uniform magnetic field whose strength is 0.8 T. If the bar magnet is oriented at an angle of $30°$ with the external field and experiences a torque of 0.2 N m, calculate:

Step 1. From $\tau=p_mB\sin\theta$: $p_m=\dfrac{\tau}{B\sin\theta}= \dfrac{0.2}{0.8\times\sin30°}=\dfrac{0.2}{0.8\times0.5}=\dfrac{0.2}{0.4}=0.5$ A m$^2$.

Step 2. The most stable configuration is $\theta=0°$ ($U_{min}=-p_mB$); the most unstable is $\theta=180°$ ($U_{max}=+p_mB$).

Step 3. Work done by an external agent moving the dipole from stable to unstable: $W=U(180°)-U(0°)=p_mB-(-p_mB)=2p_mB=2\times0.5\times0.8=0.8$ J. …

Solve for p_m from the torque equation, then use U=-p_m B cos(theta) at the two extre …

• Forgetting the factor of 2 (from PE difference between theta=0 and theta=180) when computing W. …