In a magnetic field as shown, in Fig. two horizontal wires of same mass and length `l_(1) and l_(2)` are free to slide on different vertical rails with velocities `v_(1) and v_(2)` respectively. If the resistance of two circuits are same and `a_(1) and a_(2)` are acceleration of two horizontal wires respectively, the condition for `a_(1)gta_(2)` is

Options

A. `(l_(1))/(l_(2)) = (upsilon_(2))/(upsilon_(1))`
B. `(l_(1))/(l_(2)) = ((upsilon_(2))/(upsilon_(1)))^(1//2)`
C. `(l_(1))/(l_(2)) gt ((upsilon_(2))/(upsilon_(1)))^(1//2)`
D. `(l_(1))/(l_(2)) gt ((upsilon_(2))/(upsilon_(1)))`

Answer

B

Explanation

Since `F = (B^(2) l^(2) upsilon)/(R )0`
so `(B^(2) l_(1)^(2) upsilon_(1))/(R )mg = ma_(1)`
and `(B^(2) l_(1)^(2) upsilon_(2))/(R ) = mg = ma_(2)`
Subtracting (ii) from (i)
`ma_(1) - ma_(2) = (B^(2))/(R ) (l_(1)^(2) upsilon_(1) - l_(2)^(2) upsilon_(2))`
`a_(1) - a_(2) = (B^(2))/(mR) (l_(1)^(2) upsilon_(1) - l_(2)^(2) upsilon_(2))`
For `a_(1) gt a_(2)`
`l_(1)^(2) upsilon_(1) - l_(2)^(2) upsilon_(2) gt 0` or `(l_(1))/(l_(2)) = ((upsilon_(2))/(upsilon_(1)))^(1//2)`