Question

A wire of length l is formed into a circular loop of one turn only and is suspended in a magnetic field B. When a current i is passed through the loop, the maximum torque experienced by it is

Answer 1

Given :

Length of the wire which is formed into circular loop = l

No of turns = 1

Magnetic field = B

Current passed through the loop = i

To Find :

The maximum torque experienced by the wire = ?

Solution :

Magnetic moment for the current carrying loop will be given as :

m =  $I\times A$

Here I is current in the loop and A is the area of the loop .

So, $m = \pi r^2\times I$

And here r is the radius of loop .

Since length of the wire is 'l' and it is turned into circular loop of radius 'a' , so :

$2\pi r = l$

⇒$r= \frac{l}{2\pi}$

So using these values we get :

$m = \pi \times \frac{l^2}{4\pi^2} \times I$

   $= \frac{Il^2}{4\pi}$

Now, we know that torque on a magnetic dipole of magnetic moment 'm' in uniform magnetic field 'B' is given as :

$\vec \tau =\vec m \times \vec B$

Or,$|\vec \tau | = mBsin\theta$

Here , $\theta$ is the angle between m and B

Since , for max torque $\theta$ = 90°

So,$\tau _{max}= mB$

             $= \frac{Il^2}{4\pi} \times B$

Hence , $\tau _{max}= \frac{IBl^2}{4\pi}$

So, the maximum torque experienced by the wire formed  in the circular loop is $\tau _{max}= \frac{IBl^2}{4\pi}$ .