Verified Solution • 2025–26 Session moderate

QJEEM19Ph9ae6: JEE Main 2019 Physics - Class 12 JEEM19Ph9ae6 Solution: A very long solenoid of radius R is carrying current I(t) = kte …

A very long solenoid of radius R is carrying current I(t) = kte–at(k > 0), as a function of time (t ≥\ge 0). counter clockwise current is taken to be positive. A circular conducting coil of radius 2R is placed in the equatorial plane of the solenoid and concentric with the solenoid. The current induced in the outer coil is correctly depicted, as a function of time, by :-

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/88d26426946a47c186f2a0ba8e03093c62ffbc8dec4a1ea3a5263269a2595904.webp
  • B https://ik.imagekit.io/novatra/photos/0910274c8bc1e6cccf310d3e7a409edd3a488fe166e170ef204ed9ed32d2b761.webp
  • C https://ik.imagekit.io/novatra/photos/a54c8cdcaede6a0a64feae4e4d8d3d5ac57f7713cfe898e3a82eaae60826df1c.webp
  • D https://ik.imagekit.io/novatra/photos/d0d0373cdca2708ea30dc68307caee2b72b62e1a3a2fb712fc34e4d7810fb001.webp

Step-by-Step Verified Solution

ε=−dϕdt=−ddt(μ0nl)(πR2)\varepsilon = - {{d\phi } \over {dt}} = {{ - d} \over {dt}}\left( {{\mu _0}nl} \right)\left( {\pi {R^2}} \right)

=−(μ0nπR2)dldt= - \left( {{\mu _0}n\pi {R^2}} \right){{dl} \over {dt}}

=−μ0nπR2d(kte−ωt)dt= - {\mu _0}n\pi {R^2}{{d\left( {kt{e^{ - \omega t}}} \right)} \over {dt}}

=−C(t(e−ωt)(−α)+e−ωt)(C→constant)= {\rm{ }}-C{\rm{ }}\left( {t\left( {{e^{-\omega t}}{\rm{ }}} \right){\rm{ }}\left( {-\alpha } \right){\rm{ }} + {\rm{ }}{e^{-\omega t}}{\rm{ }}} \right){\rm{ }}\left( {C{\rm{ }} \to {\rm{ }}constant} \right)

= –C (e– ω\omega t) (1 – α\alpha t)

Clearly, at t = 0

Induced current ≠\ne 0

Also, apply Lenz law to find correct option

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