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A uniform magnetic field $\vec{B}$ is perpendicular to the plane of a circular loop of diameter $10 \mathrm{~cm}$ formed from wire of diameter $2 \mathrm{~mm}$ and resistivity $2 \times 10^{-8} \Omega \mathrm{m}$. If a current of $11 \mathrm{~A}$ is to be induced in the loop then the rate at which $\vec{B}$ is to be changed is
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Verified Answer
The correct answer is:
$2.8 \mathrm{Ts}^{-}$
$$
\mathrm{i}=\frac{\mathrm{e}}{\mathrm{R}}=\frac{1}{\mathrm{R}} \frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{A}}{\mathrm{R}} \frac{\mathrm{dB}}{\mathrm{dt}}
$$
Now, $A=$ area of loop $=\pi \times(0.05)^2$
$$
=25 \pi \times 10^{-4} \mathrm{~m}^2
$$
$$
\mathrm{R}=\frac{\mathrm{\rho l}}{\text { Area of cross }- \text { section }}
$$
$$
\begin{aligned}
& =\frac{2 \times 10^{-8} \times 2 \pi \times(0.05)}{\pi \times\left(10^{-3}\right)^2} \quad[\because 1=2 \pi \mathrm{r}] \\
& =\frac{4 \times 0.05 \times 10^{-8}}{10^{-6}} \\
& =4 \times 0.05 \times 10^{-2} \\
& =2 \times 10^{-3} \Omega
\end{aligned}
$$
So, $\frac{\mathrm{dB}}{\mathrm{dt}}=\frac{\mathrm{iR}}{\mathrm{A}}=\frac{11 \times 2 \times 10^{-3}}{25 \pi \times 10^{-4}}=2.8 \mathrm{~T} / \mathrm{s}$
\mathrm{i}=\frac{\mathrm{e}}{\mathrm{R}}=\frac{1}{\mathrm{R}} \frac{\mathrm{d} \phi}{\mathrm{dt}}=\frac{\mathrm{A}}{\mathrm{R}} \frac{\mathrm{dB}}{\mathrm{dt}}
$$
Now, $A=$ area of loop $=\pi \times(0.05)^2$
$$
=25 \pi \times 10^{-4} \mathrm{~m}^2
$$
$$
\mathrm{R}=\frac{\mathrm{\rho l}}{\text { Area of cross }- \text { section }}
$$
$$
\begin{aligned}
& =\frac{2 \times 10^{-8} \times 2 \pi \times(0.05)}{\pi \times\left(10^{-3}\right)^2} \quad[\because 1=2 \pi \mathrm{r}] \\
& =\frac{4 \times 0.05 \times 10^{-8}}{10^{-6}} \\
& =4 \times 0.05 \times 10^{-2} \\
& =2 \times 10^{-3} \Omega
\end{aligned}
$$
So, $\frac{\mathrm{dB}}{\mathrm{dt}}=\frac{\mathrm{iR}}{\mathrm{A}}=\frac{11 \times 2 \times 10^{-3}}{25 \pi \times 10^{-4}}=2.8 \mathrm{~T} / \mathrm{s}$
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