A wire is bent in the form of a circular arc with a straight portion AB. Magnetic induction at O when current flowing in the wire, is

Options

A. `(mu_(0))/(2r)(pi- theta+tan theta)`
B. `(mu_(0)I)/(2pir)(pi+ theta-tan theta)`.
C. `(mu_(0)I)/(2pir)(pi- theta+tan theta)`.
D. `(mu_(0)I)/(2pir)(- tan theta +pi- theta)`.

Answer

C

Explanation

`B_(AB)=(mu_(0)I)/(4pi(OC))[2 sin theta]`
But `OC =r cos theta`
or `B_(AB)=(mu_(0)I)/(2pir) tan theta`
Magnetic field due to circular portion.
`B_(AB)'=(mu_(0)I)/(2r) (2pi- 2 theta)/(2pi)=(mu_(0)I)/(2pir)(pi-0)`
Total magnetic field
`=(mu_(0)I)/(2pir) tan theta+(mu_(0)I)/(2pir) (pi - theta)`
`=(mu_(0)I)/(2pir) [tan theta+pi - theta]`