Question

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

A. `(mu_0)/(2r) (pi- theta+tan theta)`
B. `(mu_0I)/(2piR) (pi+ theta- tan theta)`
C. `(mu_0I)/(2piR) (pi- theta+ theta)`
D. `(mu_0I)/(2piR) (-tan theta+pi- theta)`

Answer 1

Correct Answer - C
(c) `B_(AB)=(mu_0I)/(4pi(OC))[2sin theta]`
But `OC=r cos theta of B_(AB)= (mu_0I)/(2pir)tan theta`
Magnetic field due to circular potion,
`B_(AB)=(mu_0I)/(2r)((2pi- theta)/(2pi))=(mu_0I)/(2pir)(pi- theta)`
Total magnetic field
`=(mu_0I)/(2pir)tan theta+(mu_0I)/(2pir) (pi- theta)=(mu_0I)/(2pir)[tan theta+pi- theta]`