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JEE Main 202125 JulyMorning Shift
Exam
JEE Main
Year
2021
Date
25 July
Shift
Morning Shift
Subject
Physics
Chapter
Electromagnetic Induction

JEE Main 2021 · 25 July · Morning Shift · Physics · Electromagnetic Induction

A circular conducting coil of radius 1m is being heated by the...

A circular conducting coil of radius 1m1m is being heated by the change of magnetic field B→\overset{\to}{B} passing perpendicular to the plane in which the coil is laid. The resistance of the coil is 2μΩ2\mu Ω. 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.

Correct answer

80

Step-by-step solution

ϕ=B→⋅S→ϕ = \overset{\to}{B} \cdot \overset{\to}{S}

ϕ=4π×10−3(1−t100)⋅πR2ϕ = \dfrac{4}{\pi} \times {10}^{- 3}\left(1 - \dfrac{t}{100}\right) \cdot \pi{R}^{2}

ϕ=4×10−3×(1)2(1−t100)ϕ = 4 \times {10}^{- 3} \times {\left(1\right)}^{2}\left(1 - \dfrac{t}{100}\right)

ε=−dϕdtε = \dfrac{- \text{d}ϕ}{\text{d}t}

ε=−ddt(4×10−3(1−t100))ε = \dfrac{- \text{d}}{\text{d}t}\left(4 \times {10}^{- 3}\left(1 - \dfrac{t}{100}\right)\right)

ε=4×10−3(1100)=4×10−5Vε = 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=100sect = 100sec

Heat =ε2Rt= \dfrac{{ε}^{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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