Motion of Square Wire Loop Through Magnetic Field Region

To determine the correct schematic plots for the velocity $v(x)$, current $I(x)$, and force $F(x)$, we analyze the motion of the square loop of side length $L$, mass $m$, and resistance $R$ through the three distinct regions defined by the coordinate $x$ of its right edge.

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1. Region I: $0 \le x \le L$ (Loop entering the magnetic field)

• Magnetic Flux:
  As the loop enters the uniform magnetic field region ($B_0$ into the page), the length of the loop inside the field is $x$. The magnetic flux into the page is: $\Phi(x) = B_0 L x$

• Induced EMF and Current:
  By Faraday's law, the induced electromotive force is: $\mathcal{E} = -\frac{d\Phi}{dt} = -B_0 L \frac{dx}{dt} = -B_0 L v$ By Lenz's law, to oppose the increase of magnetic flux into the page, the induced current generates an outward magnetic field. Thus, the current flows in a counter-clockwise (positive) direction: $I(x) = +\frac{B_0 L v(x)}{R}$

• Magnetic Force:
  The current $I(x)$ flows upwards in the right edge of the loop. The magnetic force acting on this segment is directed to the left (retarding force): $F(x) = -I(x) L B_0 = -\frac{B_0^2 L^2}{R} v(x)$

• Equation of Motion:
  Using $m \frac{dv}{dt} = m v \frac{dv}{dx} = F(x)$: $m v \frac{dv}{dx} = -\frac{B_0^2 L^2}{R} v \implies \frac{dv}{dx} = -\frac{B_0^2 L^2}{m R} = -\alpha \quad \left(\text{where } \alpha = \frac{B_0^2 L^2}{m R} > 0\right)$

  Integrating from $x = 0$ to $x$: $v(x) = v_0 - \alpha x$

  Substituting $v(x)$ into the expression for current: $I(x) = \frac{B_0 L}{R} (v_0 - \alpha x) > 0$

  • $v(x)$ decreases linearly with $x$.
  • $I(x)$ is positive and decreases linearly as $x$ increases from $0$ to $L$.

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2. Region II: $L \le x \le 3L$ (Loop completely inside the field)

• Magnetic Flux:
  The loop is completely submerged in the magnetic field, so the magnetic flux through it is constant: $\Phi = B_0 L^2$

• Induced EMF, Current, and Force:
  Since $\frac{d\Phi}{dt} = 0$: $\mathcal{E} = 0 \implies I(x) = 0 \quad \text{and} \quad F(x) = 0$

• Velocity:
  With zero net force, the velocity remains constant: $v(x) = v(L) = v_0 - \alpha L$

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3. Region III: $3L \le x \le 4L$ (Loop exiting the magnetic field)

• Magnetic Flux:
  The length of the loop remaining inside the field is $(4L - x)$. The magnetic flux into the page is: $\Phi(x) = B_0 L (4L - x)$

• Induced EMF and Current:
  The flux into the page is decreasing: $\frac{d\Phi}{dt} = -B_0 L v$ By Lenz's law, the induced current generates an inward magnetic field to oppose this decrease. Thus, the current flows in a clockwise (negative) direction: $I(x) = -\frac{B_0 L v(x)}{R}$

• Magnetic Force:
  The current flows downwards in the left edge of the loop. The magnetic force acting on this segment is again directed to the left: $F(x) = -I(x) L B_0 = -\frac{B_0^2 L^2}{R} v(x)$

• Equation of Motion:
  $\frac{dv}{dx} = -\alpha \implies v(x) = v(3L) - \alpha (x - 3L)$

  Substituting $v(x)$ into $I(x)$: $I(x) = -\frac{B_0 L}{R} \left[ v(3L) - \alpha (x - 3L) \right]$

  • $v(x)$ continues to decrease linearly in this interval.
  • $I(x)$ is negative at $x = 3L^+$ and, as $v(x)$ decreases, $I(x)$ becomes less negative (increases linearly towards zero as $x \to 4L$).
  • $F(x)$ remains negative (retarding) throughout this region as well.

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Conclusion & Selection of Correct Options:

1. Velocity $v(x)$ [Plot (A)]: Linearly decreases in $[0, L]$, remains constant in $[L, 3L]$, and linearly decreases in $[3L, 4L]$. (Plot A is CORRECT)
2. Current $I(x)$ [Plot (B)]: Positive and linearly decreasing in $[0, L]$, zero in $[L, 3L]$, and negative moving linearly towards zero in $[3L, 4L]$. (Plot B is CORRECT)
3. Current $I(x)$ [Plot (C)]: Incorrectly shows positive current in $[3L, 4L]$. (Plot C is INCORRECT)
4. Force $F(x)$ [Plot (D)]: Incorrectly shows positive force in $[3L, 4L]$ (magnetic force always opposes motion). (Plot D is INCORRECT)

Correct Options: A and B