A conducting ring of radius 1 meter is placed in a uniform magnetic field $B$ of 0.01 tesla oscillating with frequency $100 \mathrm{~Hz}$ with its plane at right angle to $B$. What will be the induced electric field?

• A

  $\pi$ volts $/ \mathrm{m}$
• Correct answer
  B

  2 volts $/ \mathrm{m}$
• C

  10 volts $/ \mathrm{m}$
• D

  62 volts $/ \mathrm{m}$.

Solution

As a constant magnetic field conducting ring oscillates with a frequency of $100 \mathrm{~Hz}$. i.e. $T=\frac{1}{100} \mathrm{~s}$, for $\frac{T}{2}$, it goes up to $B$ $\therefore$ The corresponding frequency is $200 \mathrm{~Hz}$. $$ \begin{aligned} & \text { Induced emf }=\frac{\text { change in flux }}{\text { time }} \\ & =\frac{2 B A \cos \theta}{T}=2 B A f \cos \theta \\ & =\pi r^2 \times 2 \times 0.01 \times 200 \text { as } r=1 \mathrm{~m} \\ & =\pi 1^2 \times 2 \times 0.01 \times 200=4 \pi \mathrm{V} . \end{aligned} $$
Non-electrostatic induced electric field along the circle, $$ E=\frac{1}{2 \pi r} \times\left(\pi r^2 \times \frac{d B}{d t}\right)=\frac{e}{2 \pi r}=\frac{4 \pi}{2 \pi r}=2 \mathrm{~V} / \mathrm{m} . $$

Hence, the correct answer is option: 'B'.