A conducting loop of radius frac{10}{sqrt{pi} | vedclass

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(B) The radius of the loop is $r = \frac{10}{\sqrt{\pi}}\,cm = \frac{0.1}{\sqrt{\pi}}\,m$.The area of the loop is $A = \pi r^2 = \pi \left( \frac{0.1}

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$emf = 10\,mV$

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(B) The radius of the loop is $r = \frac{10}{\sqrt{\pi}}\,cm = \frac{0.1}{\sqrt{\pi}}\,m$.The area of the loop is $A = \pi r^2 = \pi \left( \frac{0.1}{\sqrt{\pi}} ight)^2 = \pi \left( \frac{0.01}{\pi} ight) = 0.01\,m^2$.The magnetic field changes from $B_i = 0.5\,T$ to $B_f = 0\,T$ in time $\Delta t = 0.5\,s$.The rate of change of magnetic field is $\frac{dB}{dt} = \frac{0.5 - 0}{0.5} = 1\,T/s$.According to Faraday's law,the induced emf is $\varepsilon = -\frac{d\phi}{dt} = -A \frac{dB}{dt}$.Substituting the values,$|\varepsilon| = (0.01\,m^2) 	imes (1\,T/s) = 0.01\,V$.Converting to millivolts,$|\varepsilon| = 0.01 	imes 1000\,mV = 10\,mV$.

$A$ conducting loop of radius $\frac{10}{\sqrt{\pi}}\,cm$ is placed perpendicular to a uniform magnetic field of $0.5\,T$. The magnetic field is decreased to $zero$ in $0.5\,s$ at a steady rate. The induced emf in the circular loop at $0.25\,s$ is:

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