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 $\frac{x}{\sqrt{2}}\times 10^{-4}(Nm)\hat{k}$, then find $x$.

Answer 1

Answer: 80

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