Two parallel long straight conductors lie on a smooth plane surface. Two other parallel conductors rest on them at right angles so as to form a square of side $a$ . A uniform magnetic field $B$ exists at right angles to the plane containing the conductors. Now, conductors are moving outward with a constant velocity $v_{0}$ at $t = 0$ . Then, induced current in the loop at the time $t$ is ( $λ$ is resistance per unit length of the conductors)

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$\frac{B v_{0}}{2 λ}$

$\frac{a B v_{0}}{2 λ}$

$\frac{a B v_{0}}{λ (a + v_{0} t)}$

$\frac{B v_{0}}{λ}$

answer is C.

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Detailed Solution

At $t = t$ side of square,

$\begin{array}{rcl} l & = & (a + 2 v_{0} t) \\ Area, S & = & l^{2} = {(a + 2 v_{0} t)}^{2} \\ ϕ & = & B S = B {(a + 2 v_{0} t)}^{2} \\ e & = & \frac{d ϕ}{d t} = 4 B v_{0} (a + 2 v_{0} t) \\ R & = & λ [4 l] = 4 λ (a + 2 v_{0} t) \\ ∴ i & = & \frac{e}{R} = \frac{B v_{0}}{λ} \end{array}$