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(C) The motional emf induced in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $E = B l v$.The square loop has

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$\frac{ B _{0} v _{0} d ^{2}}{ a }$

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(C) The motional emf induced in a conductor of length $l$ moving with velocity $v$ in a magnetic field $B$ is given by $E = B l v$.The square loop has two vertical sides of length $d$ at positions $x$ and $x+d$.The emf induced in the side at $x+d$ is $E _{1} = B(x+d) \cdot d \cdot v _{0} = B _{0} \left( \frac{x+d}{a} \right) d v _{0}$.The emf induced in the side at $x$ is $E _{2} = B(x) \cdot d \cdot v _{0} = B _{0} \left( \frac{x}{a} \right) d v _{0}$.The net induced emf in the loop is the difference between these two emfs: $E _{net} = E _{1} - E _{2}$.$E _{net} = \frac{B _{0} d v _{0}}{a} (x+d - x) = \frac{B _{0} v _{0} d ^{2}}{a}$.

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