Question 19: Magnetism and Matter

Question

The magnetic moment of a bar magnet is $$0.5 \mathrm{~Am}^2$$. It is suspended in a uniform magnetic field of $$8 	imes 10^{-2} \mathrm{~T}$$. The work done in rotating it from its most stable to most unstable position is:

Accepted Answer

$$\begin{aligned}
W & =\int_0^{180} d 	au \cdot d 	heta \
& =m B(\cos 0-\cos 180) \
& =0.5 	imes 8 	imes 10^{-2}(2) \
& =8 	imes 10^{-2} \mathrm{~J}
\end{aligned}
$$

Answer

$$4 	imes 10^{-2} \mathrm{~J}$$

Answer

$$16 	imes 10^{-2} \mathrm{~J}$$

Answer

$$8 	imes 10^{-2} \mathrm{~J}$$

Answer

Zero

Magnetism and Matter

The magnetic moment of a bar magnet is $0.5 \mathrm{~Am}^2. It is suspended in a uniform magnetic field of 8 \times 10^{-2} \mathrm{~T}$. The work done in rotating it from its most stable to most unstable position is:

Choose the correct answer: