Question

A thin circular ring of the area $A$is held perpendicular to a uniform magnetic field $B$of induction. 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 areas, the charge flowing through the galvanometer is:

• A. $\frac{BR}{A}$
• B. $\frac{AB}{R}$
• C. $ABR$
• D. None

Answer 1

The correct option is B

$\frac{AB}{R}$

The explanation for the correct option:

Step 1: Given data

$\text{Area}=A$

$\text{Resistance}=R$

$\text{Magnetic field}=B$

Step 2: Formula

$\phi =BA$

Step 3: Calculation

The magnetic flux is given by the formula

$\phi =BA$

$I=\frac{∆Q}{∆t}$----------(1)

The induced emf is given by:

$E={\displaystyle \frac{△\phi }{△t}}$

Here,

$∆\phi$ is the change in magnetic flux.

$∆t$ is the change in time.

After the ring is suddenly squeezed to zero areas,

$E=\frac{△\phi }{△t}$

$∆\phi ={\phi }_1-{\phi }_{2}where{\phi }_{1}isinitial=0,{\phi }_{2}isfinal={\phi }_{}$

$$\begin{gathered} =\frac{\phi -0}{t} \\ =\frac{\phi }{t}----\left(2\right) \end{gathered}$$

Induced emf is produced when the magnetic flux changes.

The current is given by
$I=\frac{E}{R}$

Here,

$I$ is the current.

$t$ is time.

$E$ is the emf.

Now substitute the value of current and induced emf in the above equation ( equating equation 1and 2)
${\displaystyle \frac{Q}{△t}}={\displaystyle \frac{BA}{R△t}}$
$Q=\frac{AB}{R}$

Hence, Option(B) is correct.