A metallic ring of radius a and resistance R | vedclass

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(C) The magnetic flux through the ring is $\phi_B = B \cdot A = (B_0 \sin \omega t)(\pi a^2)$.The induced electromotive force (emf) in the ring is $E

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the force per unit length on the ring will be proportional to $B_0^2$

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(C) The magnetic flux through the ring is $\phi_B = B \cdot A = (B_0 \sin \omega t)(\pi a^2)$.The induced electromotive force (emf) in the ring is $E = -\frac{d\phi_B}{dt} = -\frac{d}{dt}(B_0 \pi a^2 \sin \omega t) = -B_0 \pi a^2 \omega \cos \omega t$.The current in the loop is $I = \frac{E}{R} = -\frac{B_0 \pi a^2 \omega}{R} \cos \omega t$. Thus,the current oscillates with frequency $\omega$.The joule heating loss is $P = I^2 R = \frac{B_0^2 \pi^2 a^4 \omega^2}{R^2} \cos^2 \omega t \cdot R = \frac{B_0^2 \pi^2 a^4 \omega^2}{R} \cos^2 \omega t$. Thus,heat loss is proportional to $a^4$.The magnetic force on a small segment $dl$ of the ring is $dF = I(dl 	imes B)$. Since the current flows along the circumference and the magnetic field is perpendicular to the plane of the ring,the force $dF$ acts radially outward or inward. The magnitude is $dF = I B dl = (\frac{B_0 \pi a^2 \omega}{R} \cos \omega t) (B_0 \sin \omega t) dl = \frac{B_0^2 \pi a^2 \omega}{R} \sin \omega t \cos \omega t dl$.The force per unit length is $\frac{dF}{dl} = \frac{B_0^2 \pi a^2 \omega}{R} \sin \omega t \cos \omega t$,which is proportional to $B_0^2$.Due to the symmetry of the ring,the net force on the ring is zero.

$A$ metallic ring of radius $a$ and resistance $R$ is held fixed with its axis along a spatially uniform magnetic field whose magnitude is $B = B_0 \sin \omega t$. Gravity is neglected. Then,

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