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Two parallel vertical metallic rails `AB` and `CD` are separated by `1m`. They are connected at the two ends by resistances `R_1` and `R_2` as shown in the figure. A horizontal metallic bar `l` of mass `0.2 kg` slides without friction, vertically down the rails under the action of gravity. There is a uniform horizontal magnetic field of `0.6 T` perpendicular to the plane of the rails. It is observed that when the terminal velocity is attained, the powers dissipated in `R_1` and `R_2` are `0.76W` and `1.2W` respectively `(g=9.8m//s^2)`

The value of `R_1` is

Text Solution

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When rod moves downward under gravity then motional emf is generated across its ends and current starts flowing through the rod. Magnetic force thus acting on the rod tries to oppose its motion. As speed of rod increases, emf induced also increases which in turn increases the current flowing through the rod which further increases magnetic force acting on the rod. At a particular speed magnetic force becomes equal to the weight of the rod and then rod attains terminal velocity.
Hence for terminal velocity we can write the following equation:
IlB = mg
`implies I=(mg)/(lB)=(0.2 xx 9.8)/(1 xx 0.6)=(1.96)/(0.6) amp`
If e is the emf induced in the rod then el will be the power delivered by the rod as a source of emf and this will be equal to the sum of the power dissipated across both the resistors. Hence we can write the following:
`eI=P_(1)+P_(2)`
`implies e xx (1.96)/(0.6)=1.2+0.76`
`implies e xx (1.96)/(0.6)=1.96`
`implies` e = 0.6 V
Motional emf induced across the length of rod can also be written as Bvl, where v is the velocity of rod perpendicular to its length. Hence terminal velocity can be calculated as follows:
Bvl = 0.6
`implies v=(0.6)/(0.6 xx 1) = 1 m // s`
Hence rod attains a terminal velocity of `1 m // s`.
Rate of heat dissipation across a resistor when e potential difference is applied across its ends can be written as follows:
`P=(e^(2))/(R) implies R=(e^(2))/(P)`
Hence both the resistances can be calculated as follows:
`R_(1)=(e^(2))/(P_(1))=((0.6)^(2))/(1.2)=0.3 Omega`
and `R_(2)=(e^(2))/(P_(2))=((0.6)^(2))/(0.76)=0.47`
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