An air-cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500, carries a current 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.

Solution

Given, an cored solenoid.
Length of the solenoid, l = 0.30 m
Area of cross-section of solenoid, A = 25 x 10^-4m²Number of turns of the coil, N = 500
Current carried by the coil, I = 2.5 A

Initial magnetic flux, $ϕ_{1} = NBA = N (\frac{μ_{0} NI}{l}) A$
i.e., $ϕ_{1} = \frac{μ_{0} N^{2} IA}{l}$

Now, on substituting the values,
   $ϕ_{1} = 4 \times \frac{22}{7} \times 10^{- 7} \times 500 \times 500 \times 2.5 \times 25 \times 10^{- 4} \times \frac{1}{0.30}$ $ϕ_{1} = 6.55 \times 10^{- 3} Wb$

Final magnetic flux, $ϕ_{2} = 0$

Therefore, Change of flux is,
   $∆ ϕ_{B} = 0 - 6.55 \times 10^{- 3} = - 6.55 \times 10^{- 3} Wb$

Corresponding time interval, $∆ t = 10^{- 3} s$

Hence,
Average e.m.f induced across the open switch
   $= - \frac{∆ ϕ_{B}}{∆ t} = - \frac{6.55 \times 10^{- 3}}{10^{- 3}} = 6.55 V .$

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Electric Charges And Fields

Some More Questions From Electric Charges and Fields Chapter

a) An electrostatic field line is a continuous curve. That is, a field line cannot have sudden breaks. Why not?
b) Explain why two field lines never cross each other at any point?

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