Question

A rectangular current carrying coil having $100$ turns is turned in a uniform magnetic field of $\frac{0.05}{\sqrt{2}} \hat{j}$ tesla as shown in the figure. If the torque acting on the loop is $\frac{x}{\sqrt{2}} \times 10^{-4}( N m ) \hat{k}$, then find $x$.

Answer 1

The magnetic dipole moment of the current carrying coil is given by
$\vec{m} =N I A \hat{n} $
$=100 \times 0.5 \times(0.08 \times 0.04) \hat{i}$
$=16 \times 10^{-2} \,\hat{i}\, A\,m ^{2}$
The torque acting on the coil is
$\vec{\tau} =\vec{m} \times \vec{B}=m B(\hat{i} \times \hat{j}) $
$=16 \times 10^{-2} \times \frac{0.05}{\sqrt{2}} \hat{k}( N m )$