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NEET
Exam
NEET
Subject
Physics
Chapter
Magnetic Effects of Current

NEET · Physics · Magnetic Effects of Current

A very long solenoid of radius R is carrying current I(t) = kt e ^...

A very long solenoid of radius RR is carrying current I(t)=kte−\alphatI\left(t\right) = kt{e}^{- \alphat}\left(k > 0\right),asafunctionoftime(t≥0).as a function of time \left(t \ge 0\right) . Counterclockwise current is taken to be positive. A circular conducting coil of radius 2R2R is placed in the equitorial 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:A very long solenoid of radius R is carrying current I(t) = kt e ^ - (k > 0), as a function of time (t ≥ 0) .A very long solenoid of radius R is carrying current I(t) = kt e ^ - (k > 0), as a function of time (t ≥ 0) .A very long solenoid of radius R is carrying current I(t) = kt e ^ - (k > 0), as a function of time (t ≥ 0) .A very long solenoid of radius R is carrying current I(t) = kt e ^ - (k > 0), as a function of time (t ≥ 0) .

Options

  1. AA mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double slit and
  2. BA mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double slit and
  3. CA mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double slit and
  4. DA mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double slit and

Correct answer

B. A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young's double slit and

Step-by-step solution

ϕouter=(μ0nKte−\alphat)4πR2{ϕ}_{outer} = \left({\mu}_{0}nKt{e}^{- \alphat}\right)4\pi{R}^{2}
ε=−dϕdt=−μ0nK4πR2(e−\alphat[1−\alphat])ε = \dfrac{- dϕ}{dt} = - {\mu}_{0}nK4\pi{R}^{2} ( {e}^{- \alphat}\left[1 - \alphat\right] )
iinduced=−ce−\alphat[1−\alphat]Resistance{i}_{induced} = \dfrac{- c{e}^{- \alphat}\left[1 - \alphat\right]}{Resistance}
At t=0,t = 0, Iinduced=−ve{I}_{induced} = - ve
From the above equation and initial value of current we can say the correct graph is B.
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