Class 12 Physics
CBSE and UP Board
Electromagnetic Induction
Chapter-6 Exercise 6.15
Additional Exercise
NCERT Solutions for Class 12 Physics Chapter 6 Question 15
An air-cored solenoid with length 30 cm, area of cross-section 25 cm² and number of turns 500, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10⁻³s. 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.
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Length of the solenoid, 1 = 30 cm = 0.3 m
Area of cross-section, A = 25 cm2 = 25 x 10-4 m2
Number of turns on the solenoid, N = 500
Current in the solenoid, I = 2.5 A
Current flows for time, t = 10⁻3 s
Average back emf, e = dφ/dt———————Eq-1
Where, dφ = Change in flux = NAB ——————-Eq-2
B = Magnetic field strength =μ0NI/l——————–Eq-3
Where, μ0= Permeability of free space = 4π x 10-7 T m A-1
Using equations (2) and (3) in equation (1), we get
e = μ0N²IA/lt
= [(4π x 10-7) x (500)² x 2.5 x 25 x 10-4]/(0.3 x 10⁻3)
=6.5 V
Hence, the average back emf induced in the solenoid is 6.5 V.