A constant magnetic field of 1 , T is applied | vedclass

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(C) The ring has a radius $R = 1 \, m$ and moves with a constant velocity $v = 1 \, m/s$ along the $x$-axis.At $t = 0 \, s$,the center $O$ is at $x =

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$2$

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(C) The ring has a radius $R = 1 \, m$ and moves with a constant velocity $v = 1 \, m/s$ along the $x$-axis.At $t = 0 \, s$,the center $O$ is at $x = -1 \, m$. Since the radius is $1 \, m$,the rightmost edge of the ring is at $x = 0$.At $t = 1 \, s$,the center $O$ moves by a distance $d = v \cdot t = 1 \, m/s \cdot 1 \, s = 1 \, m$. Thus,the new position of the center is $x = -1 + 1 = 0 \, m$.At $t = 1 \, s$,the ring is exactly half-submerged in the magnetic field region $(x > 0)$.The induced $emf$ in a moving conductor is given by $emf = B \cdot L \cdot v$,where $L$ is the effective length of the conductor perpendicular to the velocity and the magnetic field.For a ring moving into a magnetic field,the effective length $L$ is the diameter of the chord at the boundary of the magnetic field. When the center is at $x = 0$,the chord is the vertical diameter of the ring,so $L = 2R = 2 \cdot 1 \, m = 2 \, m$.Substituting the values: $emf = 1 \, T \cdot 2 \, m \cdot 1 \, m/s = 2 \, V$.

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