Question : An air-cored solenoid with length 30 cm, area of cross-section 25 cm2 and number of turns 800, carries a current of 2.5 A. The current is suddenly switched off in a brief time of 10-3 s. Ignoring the variation in magnetic field near the ends of the solenoid, the average back emf induced across the ends of the open switch in the circuit would be
a) Zero
b) 3.125 Volts
b) 3.125 Volts
c) 6.54 Volts
d) 16.74 Volts
Doubt by Yashika
Ans : c) is the correct option.
Detailed Solution :
l = 30 cm
= 30/100 = 0.3 m
A = 25 cm2
= 25×10-4 m2
= 25×10-4 m2
N = 800
I1 = 2.5 A
I2 = 0
dI = I2-I1
dI = 0-2.5
dI = -2.5 A
dI = I2-I1
dI = 0-2.5
dI = -2.5 A
dt = 10-3 sec
dI/dt = -2.5/10-3
dI/dt = -2500 — (1)
dI/dt = -2500 — (1)
Also,
Self Inductance of the long Solenoid
Self Inductance of the long Solenoid
L = µ₀N2A/l
L= [4π×10-7(800×800)25×10-4]/0.3
L=(64π×10-5)/0.3
L=64π/3 ×10-4 H — (2)
L=(64π×10-5)/0.3
L=64π/3 ×10-4 H — (2)
Now,
Induced emf (ε)
= -LdI/dt
= -[64π/3 ×10-4]×[-2500]
= 16π/3 H
= [16×3.14]/3
= 50.24/3
= -LdI/dt
= -[64π/3 ×10-4]×[-2500]
= 16π/3 H
= [16×3.14]/3
= 50.24/3
= 16.746
Hence, d) would be the correct option.