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Q.

A rigid wire loop of square shape having side of length L  and resistance R is moving along the x-axis with a constant velocity  v0 in the plane of the paper. At t=0, the right edge of the loop enters a region of length 3L where there is a uniform magnetic field B0 into the plane of the paper, as shown in the figure. For sufficiently large v0, the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let v(x), I(x) and F(x) represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of x. Counter-clockwise current is taken as positive.Which of the following schematic plot(s) is (are) correct? (Ignore gravity)

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a

b

c

d

answer is B.

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Detailed Solution

For  O→L e=BLϑi=BLϑR F=BiL=B2L2ϑR a=B2L2ϑmR   ∵ acceleration =vdvdx       ∫v0vdv=∫0x−B2L2mRdx  V=-B2L2mRx+V0  i=−B3L3mR2x+BLRV0      Force  F=BLBLRV0−B3L3mR2xF=B2L2RV0−B4L4mR2xMagnetic force should have negative slope. From L to 3L :  enet=0i=0F=0V=constantFrom 3L to 4L: Direction of induced current will be clockwise or current flows upward in left arm of loop.The magnitude of induced current continues to decrease.So, magnetic force will act in left direction, which will again retard the motion of loop.
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