Electromagnetic Induction

JEE Main level, single correct, medium

10 of 153

Question 10 of 153

A coil is placed perpendicular to a magnetic field of 5000T5000T. When the field is changed to 3000T3000T in 2s2s, an induced emf of 22V22V is produced in the coil. If the diameter of the coil is 0.02m0.02m, then the number of turns in the coil is:

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Answer: (B) 7070

Average induced EMF is given by, ϵ=N(∆ϕt)\epsilon =N\left(\dfrac{∆\phi }{t}\right)
Also, change in flux ∆ϕ=(∆B)A∆\phi =(∆B)A
Given: Bi=5000T{B}_{i}=5000T,
Bf=3000T{B}_{f}=3000T
d=0.02m⇒r=d2=0.01md=0.02m\Rightarrow r=\dfrac{d}{2}=0.01m
Hence,
∆ϕ=(∆B)A∆\phi =(∆B)A
=(2000)π(0.01)2=0.2π=\left(2000\right)\pi {\left(0.01\right)}^{2}=0.2\pi
EMF, ϵ=N(∆ϕt)⇒22=N(0.2π2)\epsilon =N\left(\dfrac{∆\phi }{t}\right)\Rightarrow 22=N\left(\dfrac{0.2\pi }{2}\right)
⇒N=70\Rightarrow N=70

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