A conducting square frame of side $a$ and a long straight wire carrying current $I$ are located in the same pl

Question

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:

Accepted Answer

1.  **Magnetic Field:** The magnetic field $B$ at a distance $r$ from a long straight wire carrying current $I$ is given by $B = \frac{\mu_0 I}{2\pi r}$ .
2.  **Motional EMF:** A conductor of length $L$ moving with velocity $v$ perpendicular to a magnetic field $B$ induces an emf $\varepsilon = B L v$ .
3.  **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 $r_2 = x + \frac{a}{2} = \frac{2x+a}{2}$.
4.  **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: $\varepsilon_2 = \frac{\mu_0 I}{2\pi r_2} a v = \frac{\mu_0 I a v}{\pi (2x+a)}$.
5.  **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:
    $$ \varepsilon_{net} = \varepsilon_1 - \varepsilon_2 \propto \left( \frac{1}{2x-a} - \frac{1}{2x+a} 
ight) $$
    $$ \varepsilon_{net} \propto \frac{(2x+a) - (2x-a)}{(2x-a)(2x+a)} $$
    $$ \varepsilon_{net} \propto \frac{2a}{(2x-a)(2x+a)} $$
    Therefore, the induced emf is proportional to $\frac{1}{(2x-a)(2x+a)}$.

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