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
274.5k+ views
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 }}$$ = \dfrac{{4 \times 0.22}}{{2 \times \dfrac{{22}}{7}}} = 0.14m$
So, the radius of the circle is $0.14m$
The area of the square loop is ${l^2} = {\left( {0.22} \right)^2} = 0.0484{m^2}$
The area of the circle is $\pi {r^2} = \pi \times {\left( {0.14} \right)^2} = 0.06157{m^2}$
The change in area is
$
dA = \left( {0.06157 - 0.0484} \right){m^2} \\
\Rightarrow dA = 0.01317{m^2} \\
$
The magnetic flux linked with the loop is $d\Phi = BdA$
$
\Rightarrow d\Phi = 0.2 \times 0.01317 \\
\Rightarrow d\Phi = 0.002634Wb \\
$
The emf induced is given as
$\varepsilon = - \dfrac{{d\Phi }}{{dt}}$ where $\varepsilon $is the emf induced in the circle
Therefore, substituting suitable values in the above expression we get,
$
\varepsilon = - \dfrac{{0.002634}}{{0.4}} = - 0.006585V \\
\Rightarrow \varepsilon \approx - 6.6mV \\
$
Hence the correct option is B.
Note The direction of the induced emf is such that it always opposes the cause that produces it. This is called Lenz’s law. Hence in this particular problem, the emf induced opposes the change in flux which is caused due to the change in area enclosed by the wire.
Formula used
$\varepsilon = - \dfrac{{d\Phi }}{{dt}}$ where $\varepsilon $ is the emf induced, $d\Phi $ is the magnetic flux linked and $dt$ is the change in time. The negative sign indicates that the change in magnetic flux opposes the induced emf.
Complete step by step answer
We are given a square loop of side $22cm$ which is changed into a circle.
Since the square loop is changed into a circle, the perimeter of the square loop must be equal to the circumference of the circle.
Therefore we have,
$4 \times l = 2\pi r$ where $l$ is the length of the side of the square loop and $r$ is the radius of the circle.
$ \Rightarrow r = \dfrac{{4l}}{{2\pi }}$$ = \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.
Recently Updated Pages
Wheatstone Bridge – Principle, Formula, Diagram & Applications

Mass vs Weight: Key Differences Explained for Students

Circuit Switching vs Packet Switching: Key Differences Explained

Uniform Acceleration Explained: Formula, Examples & Graphs

Young’s Double Slit Experiment Derivation Explained

Classification of Drugs in Chemistry: Types, Examples & Exam Guide

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Derivation of Equation of Trajectory Explained for Students

Electron Gain Enthalpy and Electron Affinity Explained

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

Understanding Electromagnetic Waves and Their Importance

Class 12 Physics Formulas: Complete Chapter-wise PDF Guide

