Question

A coil has an area $100{cm}^{2}and50turns$. Its resistance is $2\Omega$. It is kept perpendicular to a magnetic field of induction $0.02{Wbm}^{-2}$. If it is removed in $1s$, the average induced $emf$ is

• A. $-0.01V$
• B. $0.02V$
• C. $0.03V$
• D. $0.05V$

Answer 1

The correct option is A $-0.01V$

Given: A=100 $c{m}^2$, N=50, R=2 ,b=0.02 ,t=1 sec

Solution: by using the formula

$\epsilon =-\frac{Nd\varphi }{dt}=-\frac{NBA(cos{\theta }_1-cos{\theta }_2)}{t}$

Remember that $cos{\theta }_1=0^0$(as area vector is parallel to the magnetic field) but $cos{\theta }_2$ does not have any value because it is removed (ATQ)

$\epsilon =-\frac{100\times {10}^{-4}\times 50\times 0.02}{1}=-0.01$

Hence the correct option is A