JEE Advanced 2024 · Paper 2 · Physics Q1

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PhysicsSingle correct+3 / -1Question 18 of 51

A region in the form of an equilateral triangle (in $x$ - $y$ plane) of height $L$ has a uniform magnetic field $\vec B$ pointing in the $+z$ -direction. A conducting loop PQR, in the form of an equilateral triangle of the same height $L$ , is placed in the $x$ - $y$ plane with its vertex P at $x = 0$ in the orientation shown in the figure. At $t = 0$ , the loop starts entering the region of the magnetic field with a uniform velocity $\vec v$ along the $+x$ -direction. The plane of the loop and its orientation remain unchanged throughout its motion.

Which of the following graph best depicts the variation of the induced emf ( $E$ ) in the loop as a function of the distance ( $x$ ) starting from $x = 0$ ?

Top view with axes: x points down the page, y to the right, z out of the page. The conducting loop PQR is an equilateral triangle of height L pointing down: base QR at x = −L (top) and vertex P at x = 0, moving with velocity v in the +x direction (downward). The magnetic-field region (shaded, B along +z) is an equilateral triangle of height L pointing up, with its apex at P's starting point x = 0 and its base at x = L. The two triangles meet tip to tip at x = 0 and share the same vertical axis.

E vs x: E starts at 0 at x = 0, decreases linearly to a negative minimum at x = L, then increases linearly, crossing zero at x = 3L/2 and reaching a positive maximum at x = 2L about twice the size of the minimum; at x = 2L it drops to zero (dashed vertical line) and stays zero.

E vs x: E starts at 0, decreases linearly to a negative minimum at x = L, then increases linearly, crossing zero at x = 3L/2 and reaching a positive value at x = 2L slightly smaller in size than the minimum; at x = 2L it drops to zero and stays zero.

E vs x: E starts at 0, decreases linearly to a negative minimum at x = L, then increases linearly back to zero at x = 2L (a V-shape entirely below the axis) and stays zero; L/2 and 3L/2 are marked on the axis.

E vs x: E starts at 0, decreases linearly to a negative minimum at x = L, jumps back to zero at x = L (dashed vertical line), then increases linearly from zero to a positive maximum at x = 2L, where it drops to zero and stays zero.

Answer and worked solution

Official answer: A

Let $x$ be the distance moved by P, and $\xi$ the distance below the line $x=0$ .

Widths. At depth $\xi$ :

the field triangle (apex at $x=0$ ) has width $\frac{2}{\sqrt3}\xi$ for $0\le\xi\le L$ ;
the loop (vertex at $x$ , base at $x-L$ ) has width $\frac{2}{\sqrt3}(x-\xi)$ .

Both are centred on the same axis, so the overlap width is $\frac{2}{\sqrt3}\min(\xi,\,x-\xi)$ .

0 < x < L. The overlap spans $0\le\xi\le x$ :
$A=\frac{2}{\sqrt3}\cdot\frac{x^2}{4},\qquad \frac{dA}{dx}=\frac{x}{\sqrt3}.$

L < x < 2L. The overlap spans $x-L\le\xi\le L$ :
$A=\frac{2}{\sqrt3}\left[\frac{x^2}{4}-(x-L)^2\right],\qquad \frac{dA}{dx}=\frac{2}{\sqrt3}\left(2L-\frac{3x}{2}\right).$

x > 2L. The loop has fully left the field region, so $A=0$ .

emf. $E=-Bv\,\dfrac{dA}{dx}$ :

It falls linearly from $0$ to $-\dfrac{BvL}{\sqrt3}$ at $x=L$ .
It then rises linearly, crossing zero at $x=\frac{4L}{3}$ , to $+\dfrac{2BvL}{\sqrt3}$ at $x=2L$ . That is twice the size of the minimum.
It drops to zero at $x=2L$ .

This is graph (A).

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

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JEE Advanced 2024 · Paper 2, Physics question 1