JEE Advanced 2016 · Paper 2 · Physics Q14

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PhysicsMultiple correct+4 / -2Question 14 of 54

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.

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

A square loop of side L and resistance R moving right with velocity v₀ towards a region of uniform field B₀ into the page (crosses) extending from x = 0 to x = 3L; the loop is also shown inside the field region and after leaving it. An x-axis is marked 0, L, 2L, 3L, 4L.

v(x) against x: starts at v₀ at x = 0, falls linearly to x = L, stays constant from L to 3L, then falls linearly again from 3L to 4L.

I(x) against x: positive and falling linearly from x = 0 to L, zero from L to 3L, then negative from 3L to 4L with its magnitude falling linearly (jumping to a negative value at 3L and rising towards zero).

I(x) against x: positive and falling linearly from 0 to L, zero from L to 3L, then positive again and falling linearly from 3L to 4L.

F(x) against x: negative from 0 to L (magnitude decreasing towards zero), zero from L to 3L, then positive and decreasing from 3L to 4L.

Answer and worked solution

Official answer: A, B

Entering, $0<x<L$ . The emf is $B_0Lv$ and the retarding force $\dfrac{B_0^2L^2v}{R}$ . With $mv\dfrac{dv}{dx}=-\dfrac{B_0^2L^2}Rv$ ,
$\frac{dv}{dx}=\text{constant},$
so $v$ falls linearly with $x$ .

Inside, $L<x<3L$ . The flux is constant: no current, no force, constant $v$ .

Leaving, $3L<x<4L$ . $v$ falls linearly again. That is plot (A).

Current.

Entering, the flux into the page grows, so the induced current is counter-clockwise (+), with magnitude $\propto v$ , decreasing.
Leaving, it is clockwise (−), with magnitude again decreasing.

That is plot (B); (C) has the wrong sign on exit.

Force. It always opposes the motion (negative), so (D) is false.

(A), (B)

Topics:Motional EMF Faraday's and Lenz's laws

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JEE Advanced 2016 · Paper 2, Physics question 14