Question

The pole strength of a bar magnet is 48 ampere - metre and the distance between its poles is 25 cm. The moment of the couple by which it can be placed at an angle of ${30}^{\circ }$ with the uniform magnetic intensity of flux density 0.15 Newton / ampere – metre will be

• A. None of the above
• B. 8 Newton - metre
• C. 12 Newton - metre
• D. 0.9 Newton - metre

Answer 1

The correct option is D 0.9 Newton - metre

Magnetic moment M = pole strength X magnetic length
M = 48 A-m x 25 $\times {10}^{-2}{m}^{-1}$

Magnetic torque $\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}(B=\text{magnetic field})$

magnetic torque,
$\tau =MB\sin \theta$
$=\left(48Am\right)\times 25\times {10}^{-2}\times 0.15\frac{N}{A-m}\times \sin {30}^{\circ }=0.9N-m$

External torque to be applied to keep the magnetic in equilibrium must be equal in magnitude ( and opposite in direction ) to the above magnetic torque.

Hence, option (c) is correct.