A metallic frame moves with constant velocity | vedclass

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(B) The induced $emf$ in the moving metallic frame can be modeled as two batteries connected in a loop. The side $AD$ is closer to the wire,so it expe

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$0.5\,A, ACW$

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(B) The induced $emf$ in the moving metallic frame can be modeled as two batteries connected in a loop. The side $AD$ is closer to the wire,so it experiences a stronger magnetic field,resulting in a higher induced $emf$ of $5\,V$. The side $BC$ is further away,resulting in an induced $emf$ of $2\,V$.The net $emf$ in the loop is $\varepsilon_{net} = 5\,V - 2\,V = 3\,V$.The direction of the current is determined by the larger $emf$ ($5\,V$ in $AD$),which drives the current in an anticlockwise $(ACW)$ direction.The induced current $I$ is given by Ohm's law: $I = \frac{\varepsilon_{net}}{R} = \frac{3\,V}{6\,\Omega} = 0.5\,A$.Thus,the magnitude is $0.5\,A$ and the direction is anticlockwise $(ACW)$.

$A$ metallic frame moves with constant velocity $v$ near an infinitely long current-carrying wire. At any instant,the induced $emf$ in sides $AD$ and $BC$ are $5\,V$ and $2\,V$ respectively. If the resistance of the frame is $6\,\Omega$,find the magnitude and direction of the induced current in it.

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