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GUJCET · Physics · Electromagnetic Induction
An air-cored solenoid with \(3 0 ~ c m\) length and area of cross section \(25 cm^2\) and number of turns 500 , carries a current of 2.5 A . 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?
- A 6.54 V
- B 65.4 V
- C 654 V
- D 0.654 V
Answer & Solution
Correct Answer
(A) 6.54 V
Step-by-step Solution
Detailed explanation
A
Flux \(\phi_1=\) N B A
\(\phi_1= N \left(\frac{\mu_0 NI }{l}\right) A\)
Linkage Total Flux \(\phi=\frac{\mu_0 N^2 A}{l} l\)
When the current is turned off, the magnetic flux associated with the solenoid becomes zero.
\(\therefore \phi_2=0\)
Induced emf
\(|\varepsilon|=\frac{|\Delta \phi|}{\Delta t}\)
\(\begin{array}{l}=\frac{\left|\phi_2-\phi_1\right|}{\Delta t} \\ =\frac{\mu_0 N^2 AI }{l \Delta t} \\ 4 \pi \times 10^{-7} \times(500)^2\end{array}\)
\(\begin{array}{l}\therefore|\varepsilon|=\frac{\times 25 \times 10^{-4} \times 2.5}{30 \times 10^{-2} \times 10^{-3}} \\ \therefore|\varepsilon|=6.54 V\end{array}\)
Flux \(\phi_1=\) N B A
\(\phi_1= N \left(\frac{\mu_0 NI }{l}\right) A\)
Linkage Total Flux \(\phi=\frac{\mu_0 N^2 A}{l} l\)
When the current is turned off, the magnetic flux associated with the solenoid becomes zero.
\(\therefore \phi_2=0\)
Induced emf
\(|\varepsilon|=\frac{|\Delta \phi|}{\Delta t}\)
\(\begin{array}{l}=\frac{\left|\phi_2-\phi_1\right|}{\Delta t} \\ =\frac{\mu_0 N^2 AI }{l \Delta t} \\ 4 \pi \times 10^{-7} \times(500)^2\end{array}\)
\(\begin{array}{l}\therefore|\varepsilon|=\frac{\times 25 \times 10^{-4} \times 2.5}{30 \times 10^{-2} \times 10^{-3}} \\ \therefore|\varepsilon|=6.54 V\end{array}\)
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