First slide

Magnetism

Question

The pole strength of a bar magnet is 48 Am and the distance between its poles is 25cm. The moment of the couple by which it can be placed at an angle of $30^{0}$ with the uniform magnetic field of flux density 0.15 N/Am will be.

Moderate

A

12Nm
B

18Nm
C

0.9Nm
D

0.5Nm

Solution

$M o m e n t o f t h e c o u p l e m e a n s t o r q u e a n d f l u x d e n s i t y m e a n s m a g n e t i c f i e l d .$
$∴ T o r q u e, Γ = M B S i n θ = m l B S i n θ [m = p o l e s t r e n g t h]$
$= 48 x 0. 25 x 0.15 x \frac{1}{2} = 0.9 N m$

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