The wire loop PQRS formed by joining two semi circular wires of radii `R_(1)`, and `_(2)`Ry carries a current I as shown the figure. The inagnitude of magnetic induction at the centre O is

Options

A. `(mu_(0)I)/(4)((1)/(R_(2))-(1)/(R_(1)))`
B. `(mu_(0)I)/(4)((1)/(R_(1))-(1)/(R_(2)))`
C. `mu_(0)I((1)/(R_(2))-(1)/(R_(1)))`
D. `mu_(0)I((1)/(R_(1))-(1)/(R_(2)))`

Answer

B

Explanation

The magnetic induction at O due to current through Qp and SR eill be zero .Thus the magnetic induction at O will be only due to current through semidircle of radii `R_(1)` and `R_(2)` respectively.
Magnetic field at O due to semicircular arc PS of radius `R_(1)` will be `B=(mu_(0))/(4pi)(piI)/(R_(1))` Its dirction is perpendicular to coil directed outwards . Magnetic field at O due semicircle arc RQ of radius `R_(2)` will be, `B_(2)=(mu_(0))/(4pi)(piI)/(R_(2))`.
Its direction is perpendicular to coil directed inwards . The resultant magnetic field at O,
`B=(B_(1)-B_(2))` outwards
`:. B=(mu_(0))/(4pi)Ipi[(1)/(R_(1))(1)/(R_(2))]=(mu_(0))/(4)I[(1)/(R_(1))(1)/(R_(2))]`