Electromagnetic Induction

JEE Main level, numerical answer, medium

26 of 153

Question 26 of 153

A circular conducting coil of radius 1m1m is being heated by the change of magnetic field B⃗\vec{B} passing perpendicular to the plane in which the coil is laid. The resistance of the coil is 2μΩ2\mu \Omega. The magnetic field is slowly switched off such that its magnitude changes in time as
B=4π×10−3T(1−t100)B=\dfrac{4}{\pi }\times {10}^{-3}T\left(1-\dfrac{t}{100}\right)
The energy dissipated by the coil before the magnetic field is switched off completely is E=______mJ.

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Answer: 80

ϕ=B⃗⋅S⃗\phi =\vec{B}\cdot \vec{S}
ϕ=4π×10−3(1−t100)⋅πR2\phi =\dfrac{4}{\pi }\times {10}^{-3}\left(1-\dfrac{t}{100}\right)\cdot \pi {R}^{2}
ϕ=4×10−3×(1)2(1−t100)\phi =4\times {10}^{-3}\times {\left(1\right)}^{2}\left(1-\dfrac{t}{100}\right)
ϵ=−dϕdt\epsilon =\dfrac{-\text{d}\phi }{\text{d}t}
ϵ=−ddt(4×10−3(1−t100))\epsilon =\dfrac{-\text{d}}{\text{d}t}\left(4\times {10}^{-3}\left(1-\dfrac{t}{100}\right)\right)
ϵ=4×10−3(1100)=4×10−5V\epsilon =4\times {10}^{-3}\left(\dfrac{1}{100}\right)=4\times {10}^{-5}V
When B=0B=0
1−t100=01-\dfrac{t}{100}=0
t=100sec⁡t=100\sec
Heat =ϵ2Rt=\dfrac{{\epsilon }^{2}}{R}t
Heat =(4×10−5)22×10−6×100J=\dfrac{{\left(4\times {10}^{-5}\right)}^{2}}{2\times {10}^{-6}}\times 100J
Heat =16×10−10×1002×10−6J=\dfrac{16\times {10}^{-10}\times 100}{2\times {10}^{-6}}J
Heat =0.08J=0.08J
Heat =80mJ=80mJ

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