Two thin long parallel wires seperated by a distance 'b' are carrying a current ' I' amp each . The magnitude of the force3 per unit length exerted by one wire on the other is

Options

A. `(mu_(0)i^(2))/(b^(2))`
B. `(mu_(0)i^(2))/(2pib)`
C. `(mu_(0)i)/(2pib)`
D. `(mu_(0)i)/(pib^(2))`

Answer

B

Explanation

The magnetic field due to current in wire 1 in the region of wire 2 will be `B_(1)=(mu_(0)2i)/(4pib)`
Since wire 2 having current is placed in a magnetic field `B_(1)` , it will experience a force given by `F=i(lB_(1)sin90^(@))`
`:.` Force per unit length
`(F)/(l)=ixx(mu_(0))/(4pi)x(2i)/(b)=(mu_(0)i^(2))/(2pib)[:'B=(mu_(0))/(4pi)xx(2i)/(b)]`
Hence `(b)` is the correct option.