Question

The length of solenoid is 0.3 m and the number of turns is 2000. The area of cross-section of the solenoid is $1.2\times {10}^{-3}{m}^2$. Another coil of turns 200 is wrapped over the solenoid. a current of 2 A is passed through the solenoid and its direction is changed in 0.25 sec. Then the induced emf in the coil:

• A. $4.8\times {10}^{-2}V$
• B. $4.8\times {10}^{-3}V$
• C. $3.2\times {10}^{-3}V$
• D. $3.2\times {10}^{-2}V$

Answer 1

The correct option is A $4.8\times {10}^{-2}V$

$N_1=2000$

$N_2=300$

$A=1.2\times {10}^{-3}{m}^2$

$l=0.3m$

$t=0.25s$

$I=0.2A$

we know,

Mutual inductance $\left(M\right)=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$

$\Rightarrow M=\frac{{\mu }_0{N}_1{N}_{2}A}{l}$

also,

induced emf $\left(\epsilon \right)=M.\frac{dI}{dt}$

and $\frac{dI}{dt}=\frac{2-(-2)}{0.25}=\frac{4}{0.25}=16$

so $\epsilon =M\frac{dI}{dt}=\frac{{\mu }_0{N}_1{N}_{2}A}{l}.\frac{dI}{dt}$

$\epsilon =\frac{4\pi \times {10}^{-7}\times 2000\times 300\times 1.2\times {10}^{-3}\times 16}{0.3}$

$\epsilon =0.048V=4.8\times {10}^{-2}V$ hence $\left(A\right)$ option is correct