Question

An air-cored solenoid with length $30\text{cm}$, area of cross-section $25{\text{cm}}^2$ 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.

Answer 1

Step 1: Find initial and final flux linked with the solenoid.
Formula used:$B=\frac{\left({\mu }_{0}NI\right)}{l}$
Given:
Length of the solenoid, $l=30\text{cm}=0.3m$
Area of cross section, $A=25{\text{cm}}^2=25\times {10}^{-4}{m}^2$
Number of turns in solenoid, N = 500
Current flowing through the solenoid, $I=2.5A$
As we know, Magnetic field inside solenoid $B=\frac{({\mu }_{0}NI}{l}$
Here ${\mu }_0$ = permeability of free space $=4\pi \times {10}^{-7}Tm/A$
Flux linked with solenoid, ${\varphi }_i=NBA=\frac{{\mu }_0{N}^{2}AI}{l}$
Initial flux, ${\varphi }_i=\frac{4\pi \times {10}^{-7}\times (500{)}^2\times 25\times {10}^{-4}\times 2.5}{0.3}$
${\varphi }_i=6.54\times {10}^{-3}Wb$
Final flux ${\varphi }_f=0(\text{as}I=0)$
Step 2: Find average back emf.
Given, current flows for time, $t={10}^{-3}s$
Average back emf = −(rate of change of flux)
Average back emf $=-\frac{({\varphi }_f-{\varphi }_i)}{t}$
Average back emf $=-\frac{(0-6.54\times {10}^{-3}}{{10}^{-3}}=6.54V$
Final answer : 6.54 V