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)

Question

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)

Infinity Learn · Accepted Answer

For  O→L $$e=BL$$ϑ$$i=BL$$ϑR $$F=BiL=B2L2$$ϑR $$a=B2L2$$ϑmR   ∵ 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.

Detailed Solution

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