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 `v_(0)` 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 `B_(0)` into the plane of the paper, as shown in the figure. For sufficiently large `v_(0)` 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)
A.
B.
C.
D.

Answer 1

Correct Answer - A::B
`i=(BLv)/R`-(i)[counter clockwise direction
while entering, zero when conpletely
inside and clockwise while exiting]
`F=iLB=(B^(2)L^(2)v)/R` -(ii) [Toward left while entering
and exiting and zero when completly inside]
`:. -mV(dv)/(dt)=(B^(2)L^(2)v)/R`
`:. int_(v_0)^(v) (dv)=-(B^(2)L^(2))/(mR) int_(0)^(x) dx`
`V=(V_0)=-(B^(2)L^(2))/(mR) x`
`:. V=(V_0)-(B^(2)L^(2)v)/(mR)` -(iii)
[V decreases from x=0 to x=L, remains constant for x=L to x=3L again decrease from `x=3L to x=4L]
from (i) and (iii)`
`i=(BL)/(R)[(V_0)-(B^(2)L^(2)v)/(mR)]`
[i decreases from x=0 to x=L i become zero from x=L to x=3L i change direction and decreases from x=3L to x=4L]
These characteristics are shown in graph (a) and (b) only.