Two long parallel conducting rails are placed in a uniform magnetic field. On one side, the rails are connected with a resistance `R`. Two rods `MN and M'N'` each having resistance `r` are placed as showing in Fig. Now on the rods `MN and M'N'` forces are applied such that the rods move with constant velocity `v`.

The current flowing through resustance `R` if both the rods move with the same speed `v` toward right is

Options

A. (a) `(BlR_(2)(v_(1)r_(2) - v_(2)r_(1)))/(R_(1)R_(2)(r_(1) + r_(2)) + r_(2)r_(1)(R_(1) + R_(2)))`
B. (b) `(BlR_(2)(v_(1)r_(2) + v_(2)r_(1)))/(R_(1)R_(2)(r_(1) + r_(2)) + r_(2)r_(1)(R_(1) + R_(2)))`
C. ( c) `(BlR_(2)(v_(1)r_(2) - v_(2)r_(1)))/(R_(1)R_(2)(r_(1) - r_(2)) + r_(2)r_(1)(R_(1) - R_(2)))`
D. (d) `(BlR_(2)(v_(1)r_(2) - v_(2)r_(1)))/(R_(1)R_(2)(r_(1) + r_(2)) - r_(2)r_(1)(R_(1) + R_(2)))`

Answer

A

Explanation

(ii) ( a). Case I

`I = (e)/(R + (r//2)) = (Blv)/(R + (r//2))`
Case II
` -(I_(1) - I')R - I_(1)r + e = 0` For loop (1) (i)
`r(I-(1) + U') = 2e` For loop (2) (ii)
Solve to get, `I_(1) = I' = (e)/(R )`
Hence current in 'R' is zero. ltbr
(ii). `e_(1) = Blv_(1), e_(2) = Blv_(2)`
For (1) `rarr` `e_(2) = (I - I_(1) + I_(2))r_(2) + I_(2)R_(2)`
For (2) `rarr` `e_(1) + e_(2) = (I - I_(1) + I_(2))r_(2) + Ir_(1)`
For (3) `rarr` `e_(1) = Ir_(1) + I_(1)R_(1)`
Solve to get `I_(1) = (BlR_(2)(v_(2)r_(2) - v_(2)r_(1)))/(r_(1)R_(2)(r_(1) + r_(2)) + r_(2)r_(1)(R_(1) + R_(2)))`