Two parallel vertical metallic rails AB and CD are separated by 1m. They are connected at two ends by resistanace `(R_1) and (R_2)` as shown in fig. A horizontal metallic bar L of mass 0.2 kg slides without froction vertically down the rails under the action of gravity. THere is a uniform horizontal magnetic field of 0.6 Tesla perpendicular to the plane of the rails. It is observed taht when the teminal velocity is attained, the power dissipated in `(R_1) and (R_2)` are 0.6 watt and 1.2 watt respectively. Find the terminal velocity of the bar L and the values of `(R_1)and (R_2)`.

Answer

A, C, D

Explanation

We can understand the direction of flow of induced currents by imaging a fctitious battery to be attached between E and F. The direction of induced current can be found with the help of Lenz's law.

P.d across parallel combination remains the same
Also, `(P_1)=e(i_1)=0.76W`
and (p_2)=ei_(2)=1.2W`
`:. (i_1)/(i_2)=(1.76)/(1.2) implies (1.76)/(1.2) (i_2)` ...(ii)
The horizontal metallic bar L moves with a terminal velocity
This means that the net force on the bar is zero.
`:. B(i_(1)-i_(2))=mg`
`implies (i_1)+(i_2)=(mg)/(Bl)=(0.2xx9.8)/(0.6xx1)=(49)/(15) amp.` ...(iii)
from (ii) and (iii)
`(1.76)/(1.2)(i_2)+(i_2)=(49)/(15)`
`implies (i_2)=2 amp. implies (i_1)=(19)/(15) amp. implies e=(0.76)/(19//15)=0.6V`
The induced emf across L due to the movement of bar L in a magnetic field
`e=B(v_T)Limplies(v_t)=(e)/(BL)=(0.6)/(0.6xx1)=1m//s`
Also, from (i),
`(R_1)=(e)/(i_1)=(0.6)/(19//15)=0.47(Omega) and (R_2)=(e)/(i_2)=(0.6)/(2)=0.3(Omega)`.