A circular loop of area $0.01\,{m^2}$ carrying a current of $10\, A$, is held perpendicular to a magnetic field of intensity $0.1\,T$. The torque acting on the loop is......$N-m$
AIPMT 1994, Easy
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Given:

$A=0.01 \mathrm{~m}^2, \mathrm{I}=10 \mathrm{~A}, \mathrm{~B}=0.1 \mathrm{~T},$

$\theta=0^{\circ}$

To find:

Torque acting on the loop, $\tau$

Solution:

Magnetic moment, $M=(I)(A)$

The torque acting on the loop is given as, $\vec{\tau}=\vec{M} \times \vec{B}$

$\tau=M B \sin \theta $

$\tau=1 A B \sin 0$

$\tau=0\;\mathrm{Nm}$

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