Question

magnetic field in a region is given by B=Boi+Boj torque on a circular loop x^2+z^2=r^2 carrying current i in anticlockwise

Answer 1

Given:

Magnetic Field region is given by B = B_(0) i + B_(0) j .

To find:

Torque acting on the circular ring.

Calculation:

First, let's calculate the magnetic moment of the circular coil.

$\boxed{ {x}^{2} + {z}^{2} = {r}^{2} }$

This represents a circle placed on the x z plane and the area vector is directed along the y axis.

Let magnetic moment be M.

$\therefore \: M = i \times (\pi {r}^{2} ) \: \: \hat{j}$

Magnetic Field Intensity is given by B.

$B = B_{0} \: \hat{i} \: + B_{0} \: \hat{j}$

Therefore torque experience by the circular wire:

$\therefore \: \vec \tau = \vec M \times \vec B$

$= > \vec{ \tau} = (i\pi {r}^{2} ) \: \hat{j} \times ( B_{0} \: \hat{i} \: + B_{0} \: \hat{j})$

$= > \vec{ \tau} = (i\pi {r}^{2} B_{0} ) \: \: ( - \hat{k})$

So, final answer is:

$\boxed{ \red{ \bold{ \vec{ \tau} = (i\pi {r}^{2} B_{0} ) \: \: ( - \hat{k})}}}$