Question

When a current carrying coil is placed in a uniform magnetic field of induction $B$, then a torque $\tau$ acts on it. If $I$ is the current, $n$ is the number of turns and $A$ is the face area of the coil and the normal to the coil makes an angle $\theta$ with $B$, Then

• A. $\tau =BInA$
• B. $\tau =BInA\sin \theta$
• C. $\tau =BInA\cos \theta$
• D. $\tau =BInA\tan \theta$

Answer 1

Answer: (B)

$\tau =BInA\sin \theta$

$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}$
$\overrightarrow{\tau }=MB\sin \theta$
$=niAB\sin \theta$ $(\because M=niA)$