Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

A conducting square frame of side $a$ and a long straight wire carrying current $I$ are located in the same plane as shown in the figure. The frame moves to the right with a constant velocity $v$ . The emf induced in the frame will be proportional to:

$\frac{1}{x^2}$

$\frac{1}{(2x-a)^2}$

$\frac{1}{(2x+a)^2}$

$\frac{1}{(2x-a)(2x+a)}$

Step-by-Step Solution

Magnetic Field: The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2\pi r}$ .
Motional EMF: A conductor of length $L$ moving with velocity $v$ perpendicular to a magnetic field $B$ induces an emf $\varepsilon = B L v$ .
Setup: The square frame of side $a$ has two vertical sides moving perpendicular to the magnetic field lines. Let $x$ be the distance of the center of the frame from the wire. The near side is at distance $r_1 = x - \frac{a}{2} = \frac{2x-a}{2}$ . The far side is at distance $r_2 = x + \frac{a}{2} = \frac{2x+a}{2}$ .
Induced EMF: EMF in near side: $\varepsilon_1 = \frac{\mu_0 I}{2\pi r_1} a v = \frac{\mu_0 I a v}{\pi (2x-a)}$ . EMF in far side: $\varepsilon_2 = \frac{\mu_0 I}{2\pi r_2} a v = \frac{\mu_0 I a v}{\pi (2x+a)}$ .
Net EMF: Since the magnetic field decreases with distance, $\varepsilon_1 > \varepsilon_2$ . The induced emfs in the two sides oppose each other in the loop loop (Lenz's Law). Thus, the net emf is: $\varepsilon_{net} = \varepsilon_1 - \varepsilon_2 \propto \left( \frac{1}{2x-a} - \frac{1}{2x+a} \right)$ $\varepsilon_{net} \propto \frac{(2x+a) - (2x-a)}{(2x-a)(2x+a)}$ $\varepsilon_{net} \propto \frac{2a}{(2x-a)(2x+a)}$ Therefore, the induced emf is proportional to $\frac{1}{(2x-a)(2x+a)}$ .

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