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JEE Main
Exam
JEE Main
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main · Physics · Electromagnetic Waves

A conducting ring of radius 1m is placed in a uniform magnetic...

A conducting ring of radius 1m1m is placed in a uniform magnetic field~BB,~of 0.01tesla0 . 01tesla oscillating with frequency 100100 Hz with its plane at right angles to B→\overset{→}{B}. What will be the induced electric field?
A conducting ring of radius 1m is placed in a uniform magnetic field~ B ,~of 0 . 01tesla oscillating with frequency 100

Options

  1. Aπvolt/m\pi volt/m
  2. B2volt/m2 volt/m
  3. C10volt/m10 volt/m
  4. D62volt/m62 volt/m

Correct answer

B. 2volt/m2 volt/m

Step-by-step solution

In a constant magnetic field, conducting ring oscillates with a frequency of 100100 Hz.

i.e.,~T=1100s\text{T} = \dfrac{1}{100}\text{s},~in time T2\dfrac{\text{T}}{2}

~

to zero.

⇒⇒~Induced emf =change in fluxtime= \dfrac{\text{change in flux}}{\text{time}}

=BAT/2=2BAT=2B×\pir2T=2×0.01×π×121/100=4πvolts= \dfrac{BA}{T / 2} = \dfrac{2BA}{T} = \dfrac{2B \times \pir^{2}}{T} = \dfrac{2 \times 0 .01 \times \pi \times {1}^{2}}{1 / 100} = 4\pi volts

The induced electric field along the circle, using Maxwell equation, is,∮E→⋅dl→=−dϕdt=AdBdt=e∮ \overset{→}{E} \cdot d\overset{→}{l} = - \dfrac{dϕ}{dt} = A\dfrac{dB}{dt} = e

⇒E=12\pir×(\pir2×dBdt)=e2\pir=4π2\pir=2Vm−1⇒ E = \dfrac{1}{2\pir} \times \left(\pir^{2} \times \dfrac{\text{d}B}{\text{d}t}\right) = \dfrac{e}{2\pir} = \dfrac{4\pi}{2\pir} = 2Vm^{- 1}

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