Question

A conducting coil having $500$ turn has cross-sectional area $0.15m^2$. A magnetic field of strength $0.2T$ linked perpendicular to this area changes to $1.0T$ in $0.4s$. The induced emf produced in the coil will be _________ volt.

• A. $15.0$
• B. $10.0$
• C. $75.0$
• D. $150.0$

Answer 1

The correct option is D $150.0$

Given,
$N=500$ turn
$A=0.15m^2$
$B_1=0.2T$
$B_2=1.0T$
$\Delta t=0.4s$
By Faraday's law of electromagnetic induction
$e=N\cdot \frac{\Delta B}{\Delta t}\cdot A$
$=500\times \frac{(B_2-B_1)}{\Delta t}\cdot A$
$=500\times \frac{(1.0-0.2)}{0.4}\times 0.15$
$=\frac{500\times 0.8\times 0.15}{0.4}$
$=500\times 2\times 0.15$
$=150.0V$
So, the induced emf produced in the coil will be $150V$.