Question

A magnet of moment $4A{m}^2$ is suspended in a uniform magnetic field of induction $5\times {10}^{-4}T$ with its axis at right angles. When the magnet is released from it's position, the kinetic energy that it gains in passing to the stable equilibrium position is :

• A. ${10}^{-4}$ J
• B. ${10}^{-3}$ J
• C. $2\times {10}^{-3}J$
• D. $5\times {10}^{-4}J$

Answer 1

Answer: (C)

$2\times {10}^{-3}J$

magnetic moment $m=4A{m}^2$
magnetic field $B=5\times {10}^{-4}T$
Potential energy of dipole placed in uniform field $\mu =-m.B$...(i)

From energy conservation principle:

$\Delta K+\Delta U=0$...(ii)
$K_2-K_1=(U_1-U_2)$
$K_1=0$, $U_1=0$

$U_2=-mB$ from equation (i)
$=-4\times 5\times {10}^{-4}$

From equation (ii)
$K_2-0=0+20\times {10}^{-4}$
$K_2=2\times {10}^{-3}J$