A conducting wire XY of mass $m$ and negligible resistance slides smoothly on two parallel conducting wire as shown in the fig. The closed circuit has a resistance R due to AC. AB and CD are perfect conductors. There is a magnetic field $B = B (+) \hat {k}$ write down the equation for the acceleration of the wire XY.

Question

 

$(a) \;\frac{l^2B}{mR} \; \frac{dB}{dt } x(t) + \frac{l^2B^2}{mR} (\frac{dx}{dt})$
$(b) \;\frac{l^2B}{mR} \; \frac{dB}{dt } x(t) - \frac{l^2B^2}{mR} (\frac{dx}{dt})$
$(c) \;\frac{-l^2B}{mR} \; \frac{dB}{dt } x(t) + \frac{l^2B^2}{mR} (\frac{dx}{dt})$
$(d) \;\frac{-l^2B}{mR} \; \frac{dB}{dt } x(t) - \frac{l^2B^2}{mR} (\frac{dx}{dt})$