First slide
Faraday's law of Electromagnetic induction
Question

A long solenoid of diameter 0.1 m has  2×104  turns per meter. At the centre of the solenoid, a coil of 100 turns and radius 0.01 m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0 A from 4 A in 0.05 s. If the resistance of the coil is 10π2Ω, the total charge flowing through the coil during this time is 

Difficult
Solution

 Given n=2×104;I=4 A

Initially I = 0 A

Bi=0 or ϕi=0

Finally, the magnetic field at the centre of the solenoid is given as

Bf=μonIBf=4π×10-7×2×104×4Bf=32π×10-3 T

Final magnetic flux through the coil is given as 

ϕf=nBA     =100×32π×10-3×π×(0.01)2 ϕf=32π2×10-5Tm2

Induced charge,

q=|Δϕ|R=ϕf-ϕiR=32π2×10-510π2=32×10-6C=32μC

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