Equal current I flows in two segments of a circular loop in the direction shown in figure. Radius of the loop is 7. The magnitude of magnetic field induction at the centre of the loop O is

Options

A. zero
B. `(mu_(0)Itheta)/(4pir)`
C. `(mu_(0)Itheta)/(2pir)(pi-theta)`
D. `(mu_(0)Itheta)/(2pir)(2pi-theta)`

Answer

C

Explanation

Magnetic field induction at O due to current through ACB is
`B_(1)=(mu_(0))/(4pi)(Itheta)/(r)`
It is acting perpendicular to the paper downwards . Magnetic field induction at O due to current through ABD is
`B_(1)=(mu_(0))/(4pi)(I)/(r)(2pi-theta)`
It is acting perpendicular to paper upward `:.` Total magnetic field at O due current loop is
`B=B_(2)-B_(1)=(mu_(0))/(4pi)(I)/(r)(2pi-theta)-(mu_(0))/(4pi)(I)/(r)theta=(mu_(0))/(4pi)(I)/(r)(pi-theta)`
acting perpendicular to paper upwards.