A conducting rod $A B$ moves parallel to the x-axis in the x-y plane. A uniform magnetic field $B$ pointing normally but out of the plane exists throughout the region. A force $F$ acts perpendicular to the rod so that the rod moves with uniform velocity $v$ .The force $F$ is given by (neglecting resistance of all the wires)

\frac{b v^{2} l^{2}}{R} e^{\frac{1}{R C}}

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\frac{v B^{2} l^{2}}{R}

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\frac{v B^{2} l^{2}}{R} ⎛ ⎜ ⎝ 1 - e^{\frac{- t}{R C}} ⎞ ⎟ ⎠

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\frac{v B^{2} l^{2}}{R} ⎛ ⎜ ⎝ 1 - e^{\frac{- 2 t}{R C}} ⎞ ⎟ ⎠

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Solution

The correct option is A $\frac{b v^{2} l^{2}}{R} e^{\frac{1}{R C}}$
induced emf= $V_{0} = B V$
$l$ =length of the rod $A P$
$\begin{matrix} S o, C u r r e n t = I C h a r g e = q = c v (1 - e^{\frac{- t}{R C}}) I = \frac{d q}{d t} = c v \times \frac{1}{R C} e^{\frac{- t}{R C}} I = \frac{V}{R} e^{\frac{- t}{R C}} \end{matrix}$
Force on the rod $A P = F = I l B$
$\begin{matrix} F = \frac{V_{0}}{R} l B e^{\frac{- t}{R C}} F = \frac{B V l}{R} l B e^{\frac{- t}{R C}} F = \frac{B V^{2} l^{2}}{R} e^{\frac{- t}{R C}} \end{matrix}$

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The correct option is A bv2l2R e1RCinduced emf=V0=BVl=length of the rod APSo,Current=ICharge=q=cv(1−e−tRC)I=dqdt=cv×1RCe−tRCI=VRe−tRCForce on the rod AP=F=IlBF=V0RlBe−tRCF=BVlRlBe−tRCF=BV2l2Re−tRC