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(D) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = B \cdot \pi r^2$,where $B$ is the magnetic field and $r$ is the radius o

Vedclass · Accepted Answer

$3.2\pi\, \mu V$

Vedclass · Answer

(D) The magnetic flux $\phi$ through the loop is given by $\phi = B \cdot A = B \cdot \pi r^2$,where $B$ is the magnetic field and $r$ is the radius of the loop.According to Faraday's law,the induced $emf$ is $\varepsilon = -\frac{d\phi}{dt}$.Since $B$ is constant,$\varepsilon = -\frac{d}{dt}(B \pi r^2) = -B \pi (2r) \frac{dr}{dt}$.Given: $B = 0.04\, T$,$r = 2\, cm = 0.02\, m$,and $\frac{dr}{dt} = -2\, mm/s = -2 	imes 10^{-3}\, m/s$ (negative because the radius is shrinking).Substituting the values:$\varepsilon = -(0.04) \cdot \pi \cdot 2 \cdot (0.02) \cdot (-2 	imes 10^{-3})$$\varepsilon = 0.04 \cdot \pi \cdot 0.04 \cdot 2 	imes 10^{-3}$$\varepsilon = 0.0032 	imes 10^{-3} \cdot \pi\, V$$\varepsilon = 3.2 	imes 10^{-6} \cdot \pi\, V = 3.2\pi\, \mu V$.

$A$ conducting circular loop is placed in a uniform magnetic field of $0.04\, T$ with its plane perpendicular to the magnetic field. The radius of the loop starts shrinking at a rate of $2\, mm/s$. The induced $emf$ in the loop when the radius is $2\, cm$ is:

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