Verified Solution • 2025–26 Session moderate

QJEEM21Phc4d4: JEE Main 2021 Physics - Class 12 JEEM21Phc4d4 Solution: An inductor coil stores 64 J of magnetic field energy and…

An inductor coil stores 64 J of magnetic field energy and dissipates energy at the rate of 640 W when a current of 8A is passed through it. If this coil is joined across an ideal battery, find the time constant of the circuit in seconds :

Multiple Choice Options

  • A 0.4
  • B 0.8
  • C 0.125
  • D 0.2

Step-by-Step Verified Solution

U=12Li2=64⇒L=2U = {1 \over 2}L{i^2} = 64 \Rightarrow L = 2

i2R=640{i^2}R = 640

R=640(8)2=10R = {{640} \over {{{(8)}^2}}} = 10

τ=LR=15=0.2\tau = {L \over R} = {1 \over 5} = 0.2

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