Question

A square loop of side $22cm$is changed to a circle in time $0.4s$ with its plane normal to a magnetic field $0.2T$. The emf induced is
A. $+ 6.6mV$
B. $- 6.6mV$
C. $+ 13.2mV$
D. $- 13.2mV$

Answer 1

Hint First equating the perimeter of the square loop with the circumference of the circle, obtain the radius of the circle. Then using these values calculate the change in area caused due to the transformation of a square loop into a circle. Now using this value, we can calculate the emf induced using a suitable formula.

Formula used
$\varepsilon = - \dfrac{{d\Phi }}{{dt}}$ where $\varepsilon$ is the emf induced, $d\Phi$ is the magnetic flux linked and $dt$ is the change in time. The negative sign indicates that the change in magnetic flux opposes the induced emf.

Complete step by step answer
We are given a square loop of side $22cm$ which is changed into a circle.
Since the square loop is changed into a circle, the perimeter of the square loop must be equal to the circumference of the circle.
Therefore we have,
$4 \times l = 2\pi r$ where $l$ is the length of the side of the square loop and $r$ is the radius of the circle.
$\Rightarrow r = \dfrac{{4l}}{{2\pi }}$ $= \dfrac{{4 \times 0.22}}{{2 \times \dfrac{{22}}{7}}} = 0.14m$
So, the radius of the circle is $0.14m$
The area of the square loop is ${l^2} = {\left( {0.22} \right)^2} = 0.0484{m^2}$
The area of the circle is $\pi {r^2} = \pi \times {\left( {0.14} \right)^2} = 0.06157{m^2}$
The change in area is
$dA = \left( {0.06157 - 0.0484} \right){m^2} \\\ \Rightarrow dA = 0.01317{m^2} \\\$
The magnetic flux linked with the loop is $d\Phi = BdA$
$\Rightarrow d\Phi = 0.2 \times 0.01317 \\\ \Rightarrow d\Phi = 0.002634Wb \\\$
The emf induced is given as
$\varepsilon = - \dfrac{{d\Phi }}{{dt}}$ where $\varepsilon$is the emf induced in the circle
Therefore, substituting suitable values in the above expression we get,
$\varepsilon = - \dfrac{{0.002634}}{{0.4}} = - 0.006585V \\\ \Rightarrow \varepsilon \approx - 6.6mV \\\$

Hence the correct option is B.

Note The direction of the induced emf is such that it always opposes the cause that produces it. This is called Lenz’s law. Hence in this particular problem, the emf induced opposes the change in flux which is caused due to the change in area enclosed by the wire.