Question

A short bar magnet placed with its axis at ${30}^o$ with a uniform external magnetic field of $0.35$ T experiences a torque of magnitude equal to $4.5\times {10}^{-2}$N m. The magnitude of magnetic moment of the given magnet is?

• A. $26$J $T^{-1}$
• B. $2.6$J $T^{-1}$
• C. $0.26$J $T^{-1}$
• D. $0.026$J $T^{-1}$

Answer 1

Answer: (C)

$0.26$J $T^{-1}$

Given:

The angle made by the magnetic field is, $\theta ={30}^o$

The magnetic field in the region is, $B=0.35T$

The torque acting on the bar magnet is $\tau =4.5\times {10}^{-2}Nm$

The torque acting on the magnet is given by:

$\tau =MBsin\theta$

$4.5\times {10}^{-2}=M\times 0.35\times \left(sin{30}^o\right)$

$⟹M=0.26J{T}^{-1}$