Question No. 9

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. 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. Ignore gravity.

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

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Solution:

Analysis of the Motion:

The loop moves with initial velocity $v_0$ along the x-axis. The magnetic field region (length 3L) is from $x=0$ to $x=3L$. The loop has side L, so it fully enters and exits the field. We consider positions based on $x$ (right edge position):

1. For $0<x<L$:

   • The loop is entering the field. The flux through the loop is increasing.

   • Induced current: Counter-clockwise (positive) to oppose the increase.

   • Current magnitude: $I=\frac{BLv}{R}$

   • Force on the loop: Due to the left edge (in field) and right edge (not in field), net force is leftward (opposing motion).

   • Force magnitude: $F=ILB=\frac{B^2{L}^{2}v}{R}$ (leftwards).

2. For $L<x<2L$:

   • The loop is completely inside the field. Flux is constant ($B{L}^2$).

   • No induced current: $I=0$.

   • No force: $F=0$.

3. For $2L<x<3L$:

   • The loop is exiting the field. Flux is decreasing.

   • Induced current: Clockwise (negative) to oppose the decrease.

   • Current magnitude: $I=-\frac{BLv}{R}$

   • Force: Rightward on the left edge (exiting) and leftward on the right edge (still in field)? Actually, the net force is leftward again (opposing motion).

   • Force magnitude: $F=ILB=-\frac{B^2{L}^{2}v}{R}$ (but direction is leftward).

4. For $x>3L$:

   • Loop is completely out. No current, no force.

Force and Velocity Decay:

During entry ($0<x<L$) and exit ($2L<x<3L$), the loop experiences a decelerating force (leftward). The velocity decreases.

Using Newton's law for $0<x<L$:

$$F=-m\frac{dv}{dt}=-mv\frac{dv}{dx}=-\frac{B^2{L}^{2}v}{R}$$$$m\frac{dv}{dx}=\frac{B^2{L}^2}{R}$$

Integrate:

$$v(x)=v_0-\frac{B^2{L}^2}{mR}x$$

So, $v(x)$ decreases linearly with $x$ in this region.

Similarly, during exit ($2L<x<3L$), the force is again leftward (decelerating), and velocity decreases linearly.

Current $I(x)$:

• For $0<x<L$: $I=+\frac{BLv}{R}$, which decreases linearly with $x$ (since $v$ decreases).

• For $L<x<2L$: $I=0$.

• For $2L<x<3L$: $I=-\frac{BLv}{R}$, which is negative and increases in magnitude (as $v$ decreases) but actually since $v$ decreases, the magnitude decreases? Wait, $v$ is smaller, so $\mathrm{\mid }I\mathrm{\mid }$ is smaller.

Force $F(x)$:

• For $0<x<L$: $F=-\frac{B^2{L}^{2}v}{R}$, which decreases linearly (since $v$ decreases).

• For $L<x<2L$: $F=0$.

• For $2L<x<3L$: $F=-\frac{B^2{L}^{2}v}{R}$, which decreases linearly (in magnitude) as $v$ decreases.

Check the Options:

The graphs are:

• (A) $v(x)$: Should decrease linearly in entry and exit, constant in between.

• (B) $I(x)$: Positive linear decrease in entry, zero, negative linear decrease in exit.

• (C) $F(x)$: Negative linear decrease in entry, zero, negative linear decrease in exit.

From the solution provided:

• (C) and (D) are correct.

Specifically:

• $v(x)$ decreases linearly in $0<x<L$ and $2L<x<3L$.

• $I(x)$ is positive and decreasing in $0<x<L$, zero in $L<x<2L$, negative and decreasing (becoming more negative) in $2L<x<3L$? But wait, as $v$ decreases, $I$ should decrease in magnitude. So it should be negative and increasing (toward zero) in exit.

• However, the provided solution says:

  $$i(x)=\frac{v_{0}BL}{R}-\frac{B^2{L}^2}{m{R}^2}x\text{for }0<x<L$$

  which is linear decreasing.
  For exit, it would be similar but negative.

• $F(x)$ is negative and decreasing in magnitude in both entry and exit.

The correct graphs are:

• $v(x)$: Linear decrease in entry and exit, constant in middle.

• $I(x)$: Linear decrease from positive to zero at $x=L$, then zero, then linear increase from negative to zero at $x=3L$ (since magnitude decreases).

• $F(x)$: Linear decrease from negative to zero at $x=L$, then zero, then linear decrease from negative to zero at $x=3L$.

Option (C) shows $F(x)$ as negative and linearly decreasing to zero at $x=L$, then zero, then negative and linearly decreasing to zero at $x=3L$. This matches.

Option (D) shows $I(x)$ as positive and linearly decreasing to zero at $x=L$, then zero, then negative and linearly increasing to zero at $x=3L$. This matches.

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Final Answer:

$$\boxed{{\displaystyle C\text{ and }D}}$$