Question

A thin circular ring of area A is held perpendicular to a uniform field of induction B. A small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is R. When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is,choose the correct option from the given options.
(A) BR/A
(B) AB/A
(C) ABR
(D) B²A/R

Answer 1

As the closed loop is placed in external magnetic field

total magnetic flux initially linked with the loop is given by

$\phi_1 = BA$

now when it is squeezed to zero area then its final flux is ZERO

so we have

$\phi_2 = 0$

now as per Faraday's law we can say

$\frac{d\phi}{dt} = EMF$

also we can say by ohm's law

EMF = i R

here

$i = \frac{dq}{dt}$

so we can say

$\frac{d\phi}{dt} = \frac{dq}{dt} * R$

integrating both sides

$\phi_1 - \phi_2 = Q * R$

plug in all value in above equation

$BA = Q* R$

so charge flow through it is given by

$Q = \frac{BA}{R}$

above is the charge flow through the loop

Answer 2

A thin circular ring is currently induced in loop and denotes clock by considering by help perpendicular to a uniform magnetic field of induction.

The AB/A denotes squeezed to zero area and carry out charge flowing through the A area perpendicular to a uniform magnetic field.

It identifies with small cut and made according to AB/A axis.