A long conducting wire carrying a current I is bent at `120^(@)` [Fig. 1.128]. The magnetic field B at a point on the right bisector of bending angle at a distance d from the bend is (`mu_(0)` is the permeability of free space)

Options

A. `(3mu_(0)I)/(2pid)`
B. `(mu_(0)I)/(2pid)`
C. `(mu_(0)I)/(sqrt3pid)`
D. `(sqrt3mu_(0)I)/(2pid)`

Explanation

`PQ=d,PR=(dsqrt3)/(2)=PS`
Magnetic field at P due to the portion QT,
`B=(mu_(0))/(4pi).(I)/((dsqrt3)/(2))(sin90^(@)+sin30^(@))`
`=(mu_(0))/(4pi).(2I)/(dsqrt3).(3)/(2)=(sqrt3mu_(0)I)/(4pid)`
Magnetic field at P due to the portion QU, `B_(2)=(sqrt3mu_(0)I)/(4pid)`


`:.` Total magnetic field at `P=B_(1)+B_(2)=(sqrt3mu_(0)I)/(2pid)`