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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?
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Verified Answer
The correct answer is:
2 volts $/ \mathrm{m}$
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,
$$
E=\frac{1}{2 \pi r} \times\left(\pi r^2 \times \frac{d B}{d t}\right)=\frac{e}{2 \pi r}=\frac{4 \pi}{2 \pi r}=2 \mathrm{~V} / \mathrm{m} .
$$
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,
$$
E=\frac{1}{2 \pi r} \times\left(\pi r^2 \times \frac{d B}{d t}\right)=\frac{e}{2 \pi r}=\frac{4 \pi}{2 \pi r}=2 \mathrm{~V} / \mathrm{m} .
$$
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