Solveeit Logo

Question

Physics Question on Electromagnetic induction

A rectangular coil of 100100 turns and size 0.1m×0.05m0.1 \,m \times 0.05\, m is placed perpendicular to a magnetic field of 0.1T0.1 \,T. If the field drops to 0.05T0.05 \,T in 0.050.05 second, the magnitude of the e.m.f. induced in the coil is

A

2\sqrt{2}

B

3\sqrt{3}

C

0.5\sqrt{0.5}

D

6\sqrt{6}

Answer

0.5\sqrt{0.5}

Explanation

Solution

Given, n=100n=100 turns, A=0.1×0.05m2A=0.1 \times 0.05\, m ^{2}
B1=0.1T,B2=0.05T,dt=0.05sB_{1}=0.1\, T , B_{2}=0.05\, T \,, dt =0.05\, s
We know that,
e=dϕdt=ddt(nBAcosθ)=nAdBcosθdte=\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,θ=0\theta=0^{\circ}
e=nAdBdt=100×0.1×0.05×(0.10.05)0.05\therefore e=\frac{n A d B}{d t}=\frac{100 \times 0.1 \times 0.05 \times(0.1-0.05)}{0.05}
e=0.5V\Rightarrow e=0.5 V