A long conducting wire carrying a current I is bent at `120^(@) `( see figure). The magnetic field B at a point P 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)/(2pi d)`
B. `(mu_(0)I)/(2pi d)`
C. `(mu_(0)I)/(sqrt3pi d)`
D. `(sqrt3 mu_(0)I)/(2pi d)`

Answer

D

Explanation

Magnetic field at P due to current through the bent wire is `B=2 [(mu_(0))/(4pi) (I)/(((sqrt3)/(2)d)) (sin 90^(@)+ sin 30^(@))] =2 [(mu_(0))/(4pi) (I)/(((sqrt3)/(2)d)) (1 + (1)/(2))]= (sqrt3mu_(0)I)/(2pi d)`