Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Q6.15 An air-cored solenoid with length $30cm$ , area of cross-section $25cm^{2}$ and number of turns $500$ , carries a current of $2.5A$ . The current is suddenly switched off in a brief time of $10^{-3}s$ . How much is the average back emf induced across the ends of the open switch in the circuit? Ignore the variation in magnetic field near the ends of the solenoid.

Answers (1)

best_answer

Given:

Length of the solenoid, $l=30cm=0.3m$

Area of the cross-section of the solenoid, $A=25cm^2=25*10^{-4}m^2$

Number of turns in the solenoid, $N=500$

Current flowing in the solenoid, $I=2.5A$

The time interval for which current flows, $\Delta t=10^{-3}s$

Now.

Initial flux:

$\phi _{initial}=NBA=N\left ( \frac{\mu _0NI}{l} \right )A=\frac{\mu _0N^2IA}{l}$

$\phi _{initial}=\frac{4\pi*10^{-7}*500^2*2.5*25*10^{-4}}{0.3}=6.55*10^{-3}Wb$

Final flux: since no current is flowing,

$\phi _{final}=0$

Now

Induced emf:

$e=\frac{d\phi}{dt}=\frac{\Delta \phi}{\Delta t}=\frac{\phi_{final}-\phi_{initial}}{\Delta t}=\frac{6.55*10^{-3}-0}{10^{-3}}=6.55V$

Hence, 6.55V of average back emf is induced.

Posted by

Pankaj Sanodiya

View full answer

Crack CUET with india's "Best Teachers"

HD Video Lectures
Unlimited Mock Tests
Faculty Support

cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14

anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens

anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question