An air-cored solenoid with length 30 ; cm,are | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Length of the solenoid,$l = 30 \; cm = 0.3 \; m$Area of cross-section,$A = 25 \; cm^{2} = 25 	imes 10^{-4} \; m^{2}$Number of turns on the soleno

Vedclass · Accepted Answer

$6.5$

Vedclass · Answer

(B) Length of the solenoid,$l = 30 \; cm = 0.3 \; m$Area of cross-section,$A = 25 \; cm^{2} = 25 	imes 10^{-4} \; m^{2}$Number of turns on the solenoid,$N = 500$Current in the solenoid,$I = 2.5 \; A$Time interval,$\Delta t = 10^{-3} \; s$The magnetic field inside the solenoid is $B = \mu_{0} \frac{N I}{l}$.The magnetic flux through each turn is $\phi = B A = \frac{\mu_{0} N I A}{l}$.The total flux linkage is $N \phi = \frac{\mu_{0} N^{2} I A}{l}$.The average induced back $emf$ is given by $e = \frac{\Delta(N \phi)}{\Delta t} = \frac{\mu_{0} N^{2} I A}{l \Delta t}$.Substituting the values:$e = \frac{4 \pi 	imes 10^{-7} 	imes (500)^{2} 	imes 2.5 	imes 25 	imes 10^{-4}}{0.3 	imes 10^{-3}}$$e = \frac{4 	imes 3.14159 	imes 10^{-7} 	imes 250000 	imes 2.5 	imes 25 	imes 10^{-4}}{0.3 	imes 10^{-3}}$$e = \frac{3.14159 	imes 10^{-7} 	imes 10^{6} 	imes 62.5 	imes 25 	imes 10^{-4}}{0.3 	imes 10^{-3}}$$e \approx 6.54 \; V \approx 6.5 \; V$.Thus,the average back $emf$ induced is $6.5 \; V$.

Similar Questions

When a current of $2 \ A$ is passed through a coil of $100$ turns,the magnetic flux associated with it is $5 \times 10^{-5} \ Wb$. Find the self-inductance of the coil.

The self-inductance of a circuit is numerically equal to

In an inductor,the current $I$ varies with time $t$ as $I = 5 \,A + 16 \,(A/s) \,t$. If the induced emf in the inductor is $5 \,mV$,the self-inductance of the inductor is .............. $\times 10^{-4} \,H$.

Which of the following is not the unit of self-inductance?

When the current through a solenoid increases at a constant rate, the induced current

Vedclass Test Series

Exam Paper Generator

Online Exam Module