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Question 6.17: A line charge λ per unit length is lodged uniformly onto the rim of a wheel of mass M and radius R. The wheel has light non-conducting spokes and is free to rotate without friction about its axis (Fig. 6.22). A uniform magnetic field extends over a circular region within the rim. It is given by, B = − B0 k (r ≤ a; a < R) = 0 (otherwise) What is the angular velocity of the wheel after the field is suddenly switched off?Class 12 - Physics - Electromagnetic Induction Electromagnetic Induction Page-232

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According to Faraday law of electromagnetic induction,
the induced emf is \epsilon=-\frac{d\phi}{dt}
thus, a relation between electric FIELD and rate of change of flux can be established.
e.g., -\int\vec{E}.\vec{dl}=-\frac{d\phi}{dt}
\vec{E} exist along circumstance of RADIUS 'a' due to change in magnetic flux.
e.g., E\int{dl}=-\frac{d(\pi a^2B)}{dt}\\E\times2\pi a = -\frac{d\pi a^2B}{dt}\\E=-\frac{a}{2}\frac{dB}{dt}

Linear charge density on rim is \lambda.so,total charge on rim is Q=2\pi a\lambda
so, F = QE =-\pi a^2\lambda\frac{dB}{dt}\\\implies F=m\frac{dv}{dt}=-\pi a^2\lambda\frac{dB}{dt}

in term of angular velocity, v=R\omega
m\frac{d}{dt}(R\omega)=-\pi a^2\lambda\frac{dB}{dt}\\mR\:d\omega=-\pi a^2\lambda.dB\\d\omega=-\frac{\pi a^2\lambda}{mR}dB
now, INTEGRATE both sides,
so,\omega=-\frac{\pi a^2\lambda B}{mR}

As direction of Angular velocity is along AXIS.
\vec{\omega}=-\frac{\lambda a^2\pi}{mR}B\hat{k}


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