A conducting square loop of side L, mass M and resistance R is moving in the XY plane with its edges parallel to the X and Y axes. The region has a uniform magnetic field, The magnetic field is zero everywhere else. [2025]

Question

A conducting square loop of side L, mass M and resistance R is moving in the XY plane with its edges parallel to the X and Y axes. The region $y \geq 0$ has a uniform magnetic field, $\vec{B} = B_{0} \hat{k} .$ The magnetic field is zero everywhere else. At time $t = 0$ , the loop starts to enter the magnetic field with an initial velocity $v_{0} \hat{j} m/s,$ as shown in the figure. Considering the quantity $K = \frac{B_{0}^{2} L^{2}}{R M}$ in appropriate units, ignoring self-inductance of the loop and gravity, which of the following statements is/are correct: [2025]

Smart Achievers · Accepted Answer

(2, 4)

$Current i = \frac{ε}{R} = \frac{B v ℓ}{R}$ $t = 0$

$\vec{F} = B (\hat{i}) (ℓ) (- \hat{j})$

$M a = - B_{0} [\frac{B_{0} v ℓ}{R}] (ℓ)$ $\Rightarrow a = - \frac{B_{0}^{2} ℓ^{2} v}{M R}$

$Given K = \frac{B_{0}^{2} ℓ^{2} V}{R M}$ $∴$ $a = - K v$

$\frac{d v}{d t} = - k v$ or, $\int_{v_{0}}^{v} \frac{d v}{d t} = \int_{0}^{t} (- k) d t$ $\Rightarrow \ln (\frac{v}{v_{0}}) = - k t$

$v = v_{0} e^{- k t}$ ...(i)

$\frac{d x}{d t} = v_{0} e^{- k t}$ $(∵ x \leq l)$

$\int_{0}^{x} d x = \int_{0}^{t} v_{0} e^{- k t} d t$ $= \frac{v_{0}}{k} (1 - e^{- k t})$

When $x = L$

$L = \frac{v_{0}}{k} (1 - e^{- k t})$

If $v_{0} = 3 k L$

$L = \frac{3 k L}{k} (1 - e^{- k t})$ $\Rightarrow \frac{1}{3} = 1 - e^{- k t}$

or $\frac{2}{3} = e^{- k t}$ $∴$ $t = \frac{1}{k} \ln (\frac{3}{2})$

Complete loop will enter at $t = \frac{1}{k} \ln (\frac{3}{2})$

So, option (4) is correct.

And as the loop is completely inside, then further no current flows and no force acts, i.e., $F = 0$

So option (2) is correct.

Solution

1 If $v_{0} = 1.5 K L$ , the loop will stop before it enters completely inside the region of magnetic field.

2 When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.

3 If $v_{0} = \frac{K L}{10}$ , the loop comes to rest at $t = (\frac{1}{K}) \ln (\frac{5}{2}) .$

4 If $v_{0} = 3 K L$ , the complete loop enters inside the region of magnetic field at time $t = (\frac{1}{K}) \ln (\frac{3}{2}) .$

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Solution

1 If v0=1.5 KL, the loop will stop before it enters completely inside the region of magnetic field.

2 When the complete loop is inside the region of magnetic field, the net force acting on the loop is zero.

3 If v0=KL10, the loop comes to rest at t=(1K)ln(52).

4 If v0=3 KL, the complete loop enters inside the region of magnetic field at time t=(1K)ln(32).

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1 If $v_{0} = 1.5 K L$ , the loop will stop before it enters completely inside the region of magnetic field.

3 If $v_{0} = \frac{K L}{10}$ , the loop comes to rest at $t = (\frac{1}{K}) \ln (\frac{5}{2}) .$

4 If $v_{0} = 3 K L$ , the complete loop enters inside the region of magnetic field at time $t = (\frac{1}{K}) \ln (\frac{3}{2}) .$