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Question

The magnetic flux density B is changing its magnitude at a constant rate dB/dt. A given mass m of copper having resistivity ρ and density d, drawn into a wire of radius a and formed into a circular loop of radius r is placed perpendicular to the field B. The induced current in the loop i is :

A
m2πρd dBdt
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B
m4πa2r dBdt
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C
m4πad dBdt
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D
m4πρd dBdt
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Solution

The correct option is D m4πρd dBdt
Area of loop, A=πr2

Magnetic flux passing through coil is

ϕ=BAcosθ=BAcos0=B(πr2)

From Faraday's law of induction

E=dϕdt=ddtB(πr2)

E=πr2dBdt

Length of the loop, l=2πr
and area of cross-section of wire, A=πa2

So, resistance of the wire will be

R=ρlA=ρ(2πr)πa2

Induced current in the loop will be

i=ER=πr2dBdtρ(2πr)πa2=πa2r2ρdBdt .........(1)

Further mass of wire is,

m=(πa2)(2πr)(d)

πa2r=m2πd

Putting in equation (1), we get

i=m4πρddBdt

Hence, option (D) is correct.

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