Two long parallel horizontal rails,a distance $l$ apart,each having a resistance $\lambda$ per unit length,are joined at one end by a resistance $R$. $A$ perfectly conducting rod $MN$ of mass $m$ is free to slide along the rails without friction. There is a uniform magnetic field of induction $B$ normal to the plane of the paper and directed into the paper. $A$ variable force $F$ is applied to the rod $MN$ such that,as the rod moves,a constant current $i$ flows through the circuit. The applied force $F$ as a function of distance $x$ of the rod from $R$ is:

  • A
    $i l B + \frac{2m \lambda i^2}{B^2 l^2}(R + 2 \lambda x)$
  • B
    $i l B + \frac{4m \lambda i^2}{B^2 l^2}(R + 2 \lambda x)$
  • C
    $i l B - \frac{2m \lambda i^2}{B^2 l^2}(R + 2 \lambda x)$
  • D
    none of these

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$A$ conducting wire $XY$ of mass $m$ and negligible resistance slides smoothly on two parallel conducting wires as shown in the figure. The closed circuit has a resistance $R$ due to $AC$. $AB$ and $CD$ are perfect conductors. There is a magnetic field $\vec{B} = B(t) \hat{k}$.
$(i)$ Write down the equation for the acceleration of the wire $XY$.
$(ii)$ If $\vec{B}$ is independent of time,obtain $v(t)$,assuming $v(0) = u_0$.
$(iii)$ For $(ii)$,show that the decrease in kinetic energy of $XY$ equals the heat lost in $R$.

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