Question:

A rectangular coil of $100$ turns and size $0.1 \,m \times 0.05\, m$ is placed perpendicular to a magnetic field of $0.1 \,T$. If the field drops to $0.05 \,T$ in $0.05$ second, the magnitude of the e.m.f. induced in the coil is

Updated On: Apr 16, 2024
  • $\sqrt{2}$
  • $\sqrt{3}$
  • $\sqrt{0.5}$
  • $\sqrt{6}$
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is C

Solution and Explanation

Given, $n=100$ turns, $A=0.1 \times 0.05\, m ^{2}$
$B_{1}=0.1\, T , B_{2}=0.05\, T \,, dt =0.05\, s$
We know that,
$e=\left|\frac{-d \phi}{d t}\right|=\frac{d}{d t}(n B A \cos \theta)=\frac{n A d B \cos \theta}{d t} $
Here,$\theta=0^{\circ}$
$\therefore e=\frac{n A d B}{d t}=\frac{100 \times 0.1 \times 0.05 \times(0.1-0.05)}{0.05}$
$\Rightarrow e=0.5 V$
Was this answer helpful?
1
0

Top Questions on Electromagnetic induction

View More Questions

Concepts Used:

Electromagnetic Induction

Electromagnetic Induction is a current produced by the voltage production due to a changing magnetic field. This happens in one of the two conditions:-

  1. When we place the conductor in a changing magnetic field.
  2. When the conductor constantly moves in a stationary field.

Formula:

The electromagnetic induction is mathematically represented as:-

e=N × d∅.dt

Where

  • e = induced voltage
  • N = number of turns in the coil
  • Φ = Magnetic flux (This is the amount of magnetic field present on the surface)
  • t = time

Applications of Electromagnetic Induction

  1. Electromagnetic induction in AC generator
  2. Electrical Transformers
  3. Magnetic Flow Meter