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 \(\mathrm{v}_{0}\) in the plane of the paper. At \(t=0\), the right edge of the loop enters a region of length \(3 L\) where there is a uniform magnetic field \(B_{0}\) into the plane of the paper, as shown in the figure. For sufficiently large \(\mathrm{v}_{0}\), the loop eventually crosses the region. Let \(x\) be the location of the right edge of the loop. Let \(\mathrm{v}(x), I(x)\) 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. Which of the following schematic plot(s) is(are) correct? (Ignore gravity)

(A) (see structure); (B) (see structure)

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Electromagnetic Induction

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

3 L

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.

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

(A)(see structure)
✓ Correct
(B)(see structure)
✓ Correct
(C)(see structure)
(D)(see structure)

Answer: (A), (B)

Solution

Region 1 (0 < x < L): Right edge entering field.

Induced emf:

ε = B Lv

, current

i = \frac{B Lv}{R}

(positive/counter-clockwise)

Force:

F = B i L = \frac{B ^{2} L ^{2} v}{R}

(leftward, opposing motion)

Equation of motion:

m \frac{d v}{d t} = - \frac{B ^{2} L ^{2} v}{R}

This gives

v (x) = v_{0} - \frac{B ^{2} L ^{2}}{m R} x

(linear decrease)

Region 2 (L < x < 3L): Entire loop in field, no flux change,

i = 0

F = 0

v

constant.

Region 3 (3L < x < 4L): Left edge leaving field.

Current reverses (clockwise/negative), force opposes motion again.

Plots (A) and (B) correctly show these features: the current is positive and linearly decreasing while the loop enters, zero while it is fully inside, and negative while it leaves, with the force a mirror of the current; the velocity falls linearly, stays constant, then falls linearly again.

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · One or more correct options

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JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Electromagnetic Induction

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