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(C) $1$. The side $PS$ is outside the magnetic field region. Since the magnetic field $B$ at the location of side $PS$ is zero,the induced $e.m.f.$ in

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Zero,$\frac{Fr}{6Bl}$

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(C) $1$. The side $PS$ is outside the magnetic field region. Since the magnetic field $B$ at the location of side $PS$ is zero,the induced $e.m.f.$ in side $PS$ is $\varepsilon = B \ell v = 0 	imes \ell 	imes v = 0$.$2$. The loop is being pulled with a constant speed $v$ by force $F$. The magnetic force on the side $QR$ (which is inside the field) is $F_m = B I \ell$. Since the velocity is constant,$F = F_m = B I \ell$.$3$. The induced current $I$ in the loop is $I = \frac{\varepsilon_{total}}{r} = \frac{B \ell v}{r}$.$4$. Substituting $I$ into the force equation: $F = B \ell \left( \frac{B \ell v}{r} ight) = \frac{B^2 \ell^2 v}{r}$,which gives $v = \frac{Fr}{B^2 \ell^2}$.$5$. The potential difference between $P$ and $S$ is the voltage drop across the resistance of that segment. The total resistance of the loop is $r$. The side $PS$ has length $\ell$. Assuming the loop is made of uniform wire,the resistance of side $PS$ is $r_{PS} = \frac{r}{6}$ (as the total perimeter is $2(2\ell + \ell) = 6\ell$).$6$. Thus,$V_{PS} = I \cdot r_{PS} = \left( \frac{B \ell v}{r} ight) \cdot \frac{r}{6} = \frac{B \ell}{6} \cdot \left( \frac{Fr}{B^2 \ell^2} ight) = \frac{Fr}{6 B \ell}$.

$A$ rectangular loop $PQRS$ is being pulled with a constant speed into a uniform transverse magnetic field by a force $F$ (as shown). The $e.m.f.$ induced in side $PS$ and the potential difference between points $P$ and $S$ respectively are (Resistance of the loop $= r$)

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