A rigid wire loop of square shape having side of length L and resistance R is moving along the -axis with a constant velocity in the plane of the paper. At , the right edge of the loop enters a region of length 3L where there is a uniform magnetic field

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. [2016]

Which of the following schematic plot(s) is (are) correct

(Ignore gravity)

Which of the following schematic plot(s) is (are) correct?

(Ignore gravity)

Smart Achievers · Accepted Answer

(1, 2)

$i = \frac{e}{R} = \frac{B L v}{R}$ ...(i)

[Counter-clockwise direction while entering, zero when completely inside and clockwise while exiting]

$F = i L B = \frac{B^{2} L^{2} v}{R}$ ...(ii)

[Toward left while entering and exiting and zero when completely inside]

$∴$ $- m v \frac{d v}{d x} = \frac{B^{2} L^{2} v}{R}$

$∴$ $\int_{v_{0}}^{v} d V = - \frac{B^{2} L^{2}}{m R} \int_{0}^{x} d x$ $\Rightarrow V - V_{0} = - \frac{B^{2} L^{2}}{m R} x$

$∴$ $V = V_{0} - \frac{B^{2} L^{2} x}{m R}$ ...(iii)

[V decreases from $x = 0$ to $x = L$ , remains constant for $x = L$ to $x = 3 L$ , again decreases from $x = 3 L$ to $x = 4 L$ hence graph (1) is correct]

From (i) and (iii),

$i = \frac{B L}{R} [V_{0} - \frac{B^{2} L^{2} x}{m R}]$

[ $i$ decreases from $x = 0$ to $x = L$ , $i$ becomes zero from $x = L$ to $x = 3 L$ , $i$ changes direction and decreases from $x = 3 L$ to $x = 4 L$ ]

Hence graph (2) is correct.

Solution

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