A uniform magnetic field B that is perpendicu | vedclass

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(B) According to Faraday's law of electromagnetic induction,the induced emf $\varepsilon$ in a loop is given by $\varepsilon = -\frac{d\Phi_B}{dt}$,wh

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$\varepsilon$

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(B) According to Faraday's law of electromagnetic induction,the induced emf $\varepsilon$ in a loop is given by $\varepsilon = -\frac{d\Phi_B}{dt}$,where $\Phi_B$ is the magnetic flux through the loop.The magnetic flux $\Phi_B$ through a loop is given by $\Phi_B = \int B \cdot dA$. Since the magnetic field $B$ is confined only to the region of radius $a$,the flux through any loop of radius $r \ge a$ is simply the flux through the region of radius $a$,which is $\Phi_B = B \cdot (\pi a^2)$.For the loop of radius $b$ (where $b > a$),the flux is $\Phi_b = B \pi a^2$. The induced emf is $\varepsilon = -\frac{d}{dt}(B \pi a^2) = -\pi a^2 \frac{dB}{dt}$.For the loop of radius $2b$ (where $2b > a$),the flux is $\Phi_{2b} = B \pi a^2$. The induced emf is $\varepsilon_{2b} = -\frac{d}{dt}(B \pi a^2) = -\pi a^2 \frac{dB}{dt}$.Since the magnetic flux linked with both loops is identical,the induced emf in both loops is the same. Therefore,the induced emf in the loop of radius $2b$ is $\varepsilon$.

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