A current carrying circular loop is perpendic | vedclass

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(B) Given: Magnetic field $B = 10^{-4} \, T$, radius $r = 20 \, cm = 0.2 \, m$.The rate of change of radius is $\frac{dr}{dt} = -2 \, mm/s = -2 	imes

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$0.08$

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(B) Given: Magnetic field $B = 10^{-4} \, T$, radius $r = 20 \, cm = 0.2 \, m$.The rate of change of radius is $\frac{dr}{dt} = -2 \, mm/s = -2 	imes 10^{-3} \, m/s$.The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = B \cdot \pi r^2$.According to Faraday's law, the induced emf $\varepsilon$ is $\varepsilon = -\frac{d\phi}{dt}$.Substituting the expression for $\phi$: $\varepsilon = -\frac{d}{dt}(B \cdot \pi r^2) = -B \pi \cdot 2r \cdot \frac{dr}{dt}$.Substituting the values: $\varepsilon = -(10^{-4}) \cdot \pi \cdot 2 \cdot (0.2) \cdot (-2 	imes 10^{-3})$.$\varepsilon = 10^{-4} \cdot \pi \cdot 0.4 \cdot 2 	imes 10^{-3} = 0.8 \pi 	imes 10^{-7} \, V$.Converting to microvolts: $\varepsilon = 0.08 \pi 	imes 10^{-6} \, V = 0.08 \pi \, \mu V$.

$A$ current carrying circular loop is perpendicular to a magnetic field of induction $10^{-4} \, T$. If the radius of the loop starts shrinking at a uniform rate of $2 \, mm/s$, then the emf induced in the loop at the instant, when its radius is $20 \, cm$ will be (in $\pi \, \mu V$)

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