Question

A current of 1.5 A flows through a solenoid of length 20.0 cm, cross-section 20.0 ${cm}^2$ and 400 turns. The current is suddenly switched off in a short time of 1.0 milisecond. Ignoring the variation in the magnetic field near the ends, the average back emf inducted in the solenoid is:

• A. 0.3 V
• B. 9.6 V
• C. 30.0 V
• D. 3.0 V

Answer 1

The correct option is D 3.0 V

Given :

$I=1.5A$

$l=0.2m$

$N=400$

$A=20\times {10}^{-4}{m}^2$

$dt={10}^{-3}s$

Solution :

We know that magnetic filed inside a solenoid is ${\mu }_{0}nI$

where $n\text{(number of turns per length)}=\frac{N}{L}=2000$

Also, the electromagnetic induction is defined as changed in flux.

So, $\text{emf}=\frac{d\varphi }{dt}$ , where $\varphi =\int B\cdot da$

$d\varphi =BNA$ $(\because \text{Total Area of Solenoid is number of turns times the area of one cross section})$

So, $\text{emf}=\frac{{\mu }_{0}nINA}{dt}$

$\text{emf}=\frac{(4\pi \times {10}^{-7})\cdot 2000\cdot 1.5\cdot 400\cdot (20\times {10}^{-4})}{{10}^{-3}}$

$\text{emf}=3.0$

The correct Opt : {D}