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 sufficient large $v_0$, the loop eventually crosses the region. Let $x$ be the location of the right edge of the loop. Let $v\left(x\right),I\left(x\right)$ and $F\left(x\right)$ 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

Answer: (B), (C)

When the loop is entering inside the magnetic field ( $x<L$) inside flux is increasing, so the current will be in counter-clockwise direction and according to lenz's law the force will be in $-$ve $x$ direction in the whole journey and the velocity decreases while entering and leaving the magnetic field.



At any moment force will be$F=BiL=\frac{v{B}^2{L}^2}{R},i=\frac{vBL}{R}\frac{F}{m}=\frac{v{B}^2{L}^2}{mR}=Kv=-v\frac{dv}{dx}{\int }_{v_0}^{v}dv=-K{\int }_0^{x}dxv=v_0-kxf=\frac{B^2{L}^2}{R}(v_0-kx)=\alpha -\beta xI=(v_0-kx)\frac{BL}{R}=i_0-\gamma x$

For $3L>x>L,f=0,I=0$ velocity is constant in this region.

$F=BiL=\frac{v{B}^2{L}^2}{R}a=\frac{B^2{L}^2}{mR}v=kv=-v\frac{dv}{dx}v={v^{\prime }}_0-kxf={\alpha }^{\prime }-{\beta }^{\prime }xI={i^{\prime }}_0-{\gamma }^{\prime }x$

Here option D is wrong as the current is changing its direction while its leaving the magnetic field region.
And from these equations, options B and C are correct.