Question

A circular ring of radius 1 m is placed in a time varying magnetic field, dB /dt =2 T/s. The potential difference between P and Q is (where angle POQ= 90)

Answer 1

Given :

Radius of the circular ring =1 m

The time varying magnetic field is  given as :

$=\frac{dB}{dt}= 2 \frac{T}{s}$

To Find :

If angle POQ = 90 , then potential difference between P and Q is = ?

Solution :

According the Faraday's law induced emf  is given as :

$E= -\frac{d\phi_B}{dt}$   -(1)

And,

$\Phi_B= B.A$

Where $\Phi_B$ is magnetic flux , B is magnetic field and A is area

So, we can also write  :

$\Phi_B = B .\pi r^2$

Differentiating it w.r.t  time we get :

$\frac{d\Phi_B}{dt} =\pi r^2\frac{dB}{dt}$

      = $\pi \times 1^2 \times 2$

      $=2 \pi\\= 6.28$

Now since the  angle between P and Q is 90 so , potential difference between P and Q is = $\frac{1}{4} \times 6.28$ V

                                 = 1.57 V

So, the potential difference between P and Q is 1.57 V.