Two long, parallel conductors carry currents in the same direction, as shown in figure. Conductors A held firmly in position. Conductors B carries a current `I_(B)` and is allowed to slide freely up and down (parallel to A) between a se of non-conducting guides. the mass per unit length of conductors B is 0.1 g/cm and the distance between the two conductors is 5 cm. if system of conductors is in equilibrium , the value of current `I_(B)` is

Options

A. 250 A
B. 240 A
C. 220 A
D. 230 A

Answer

A

Explanation

When system of conductors is in equilibrium,
The magnetic force of attraction per unit length between conductors = weight of conductor B per unit length.
`(mu_(0))/(2pi)(I_(A)xxI_(B))/(d)=(mg)/(L)=((m)/(L))g`
`(mu_(0))/(2pi)(I_(A)xxI_(B))/(d)=((m)/(L))g`
`2xx10^(-7)xx(10xxI_(B))/(0.05)=(0.01kg//m)xx10`
On solving `I_(B)=250A`