Three resistances of magniute R each are connected in the from of an equilateral traingle of side a . The combinates is placed in a magnetic field `B= B_(0)e^(-lambdat)` perpendicular to the plane . The induced current in the circult is given by.

Options

A. `((a^(2)lambda)/(2 sqrt(3R))B_(0))e^(-lambdat)`
B. `((a^(2)lambda)/(4 sqrt((3)R))B_(0))e^(-lambdat)`
C. `((a^(2)lambda)/(lambda4 (sqrt(3)R))B_(0))e^(-lambdat)`
D. `((a^(2)B_(0)R)/(lambda4 (sqrt(3))B_(0)))e^(-lambdat)`

Answer

B

Explanation

`phi = BA = B_(0)e^(-lambda t) (sqrt3)/(4) a^(2)`
`e = -(dphi)/("dt") = -(sqrt3)/(4)a^(2)B_(0)(d)/("dt")(e^(-lambdat)) = -(sqrt3)/(4)a^(2)B_(0)(-lambda)e^(-lambda t) = (sqrt3)/(4)a^(2)B_(0)lambda e^(-lambda t)`
`therefore " " l = (e)/(3R) = (sqrt3)/(12R) a^(2)B_(0) lambda e^(-lambda t) = ((a^(2)lambdaB_(0))/(4sqrt3 R))e^(-lambda t)`