Current I is flowing in a conductor shaped as shown in the figure . The radius of the curved part is r and the length of straight portion is very large . The value of the magnetic field at the centre O will be

Options

A. `(mu_(0)I)/(4pir)((pi)/(2)+1)`
B. `(mu_(0)I)/(4pir)((pi)/(2)-1)`
C. `(mu_(0)I)/(4pir)((3pi)/(2)+1)`
D. `(mu_(0)I)/(4pir)((3pi)/(2)-1)`

Answer

C

Explanation

For the circular part ABC, the angle subtended at the centre is `3pi//2`.

Due to ABC , `B_(1)=(mu_(0))/(4pi)*(I)/(r)((3pi)/(2))`
Due to AD , A is at the end of the wire , therefore at O,
`B_(2)=(mu_(0)I)/(2pir)xx(1)/(2)=(mu_(0)I)/(4pir)`.
`therefore` Total induction `=(mu_(0)I)/(4pir){(3pi)/(2)+1}`.