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(B) According to Faraday's Law of induction,the changing magnetic flux induces an electric field $E$ along the circumference of the wheel. The induced

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$\frac{\pi a b^2 \lambda B}{I}$ clockwise as seen from above

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(B) According to Faraday's Law of induction,the changing magnetic flux induces an electric field $E$ along the circumference of the wheel. The induced $EMF$ $\varepsilon$ is given by $\varepsilon = \left| -\frac{d\phi}{dt} ight| = \oint \vec{E} \cdot d\vec{l} = E(2\pi a)$.Since $\phi = B(\pi b^2)$,we have $\varepsilon = \pi b^2 \frac{dB}{dt}$.Equating the two,$E(2\pi a) = \pi b^2 \frac{dB}{dt}$,which gives $E = \frac{b^2}{2a} \frac{dB}{dt}$.The force on the charge element $dq = \lambda(2\pi a)$ is $dF = E dq$. The torque $	au$ is $	au = a F = a(E \cdot 2\pi a \lambda) = 2\pi a^2 \lambda E$.Substituting $E$,$	au = 2\pi a^2 \lambda \left( \frac{b^2}{2a} \frac{dB}{dt} ight) = \pi a b^2 \lambda \frac{dB}{dt}$.Using $	au = I \frac{d\omega}{dt}$,we have $I \frac{d\omega}{dt} = \pi a b^2 \lambda \frac{dB}{dt}$.Integrating both sides,$I \omega = \pi a b^2 \lambda B$,so $\omega = \frac{\pi a b^2 \lambda B}{I}$.By Lenz's Law,the induced current creates a magnetic field opposing the change. Since the magnetic field is decreasing,the induced electric field will create a torque such that the wheel rotates in the direction that opposes the decrease in flux,which is clockwise as seen from above.

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