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A metal loop of area $10 \mathrm{~cm}^2$ is placed in a region such that its area vector points along $\hat{\mathrm{k}}$. The region contains a uniform magnetic field of magnitude $1.73 \mathrm{~T}$ that points in the direction $\hat{i}+\hat{j}+\hat{k}$. When the magnetic field is switched off, the field decreases to zero at a steady rate in $10 \mathrm{~s}$, then the magnitude of emf induced in the loop is
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The correct answer is:
$0.10 \mathrm{mV}$
We have
$\mathrm{e}=-\frac{\Delta \phi}{\Delta \mathrm{t}}$
$\Rightarrow \quad \mathrm{e}=\frac{-\left(\phi_{\mathrm{f}}-\phi_{\mathrm{i}}\right)}{\Delta \mathrm{t}}=\frac{\phi_{\mathrm{i}}-\phi_{\mathrm{f}}}{\Delta \mathrm{t}}$
$=\frac{1.73\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\right) \cdot\left(10 \times 10^{-4}\right) \hat{\mathrm{k}}-0}{10}$
$=\frac{\frac{1.73}{\sqrt{3}} \times 10^{-3}}{10}=10^{-4}$ volt $=0.1 \mathrm{mV}$
$\mathrm{e}=-\frac{\Delta \phi}{\Delta \mathrm{t}}$
$\Rightarrow \quad \mathrm{e}=\frac{-\left(\phi_{\mathrm{f}}-\phi_{\mathrm{i}}\right)}{\Delta \mathrm{t}}=\frac{\phi_{\mathrm{i}}-\phi_{\mathrm{f}}}{\Delta \mathrm{t}}$
$=\frac{1.73\left(\frac{\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}}{\sqrt{3}}\right) \cdot\left(10 \times 10^{-4}\right) \hat{\mathrm{k}}-0}{10}$
$=\frac{\frac{1.73}{\sqrt{3}} \times 10^{-3}}{10}=10^{-4}$ volt $=0.1 \mathrm{mV}$
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