Search any question & find its solution
Question:
Answered & Verified by Expert
A coil having 500 square loops each of side 10 $\mathrm{cm}$ is placed normal to a magnetic flux which increases at a rate of $1 \mathrm{Ts}^{-1}$. The induced emf is
Options:
Solution:
2987 Upvotes
Verified Answer
The correct answer is:
$5 \mathrm{~V}$
Given that, number of loops, $N=500$
Side of square, $a=10 \mathrm{~cm}=0.1 \mathrm{~m}$
Rate of increase of magnetic field, $\frac{d B}{d t}=1 \mathrm{~T} / \mathrm{s}$
Now, induced emf, $\varepsilon=-N \frac{d \phi}{d t}=-N \frac{d}{d t}(B A)$
$=-N A \frac{d B}{d t}=-N a^2 \frac{d B}{d t}$ $\left[\because\right.$ Area of square $\left.A=a^2\right]$
Substituting the values, we get
$\varepsilon=-500 \times(0.1)^2 \times 1=-5 \mathrm{~V}$
Magnitude, $|\varepsilon|=5 \mathrm{~V}$
Side of square, $a=10 \mathrm{~cm}=0.1 \mathrm{~m}$
Rate of increase of magnetic field, $\frac{d B}{d t}=1 \mathrm{~T} / \mathrm{s}$
Now, induced emf, $\varepsilon=-N \frac{d \phi}{d t}=-N \frac{d}{d t}(B A)$
$=-N A \frac{d B}{d t}=-N a^2 \frac{d B}{d t}$ $\left[\because\right.$ Area of square $\left.A=a^2\right]$
Substituting the values, we get
$\varepsilon=-500 \times(0.1)^2 \times 1=-5 \mathrm{~V}$
Magnitude, $|\varepsilon|=5 \mathrm{~V}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.