The wire loop PQRSP formed by joining two semicircular wires of radii R1 and R2 carries a current I as shown. The magnitude of the magnetic induction at the centre C is ____________.

Answer

`(mu_(0)I)/4(1/(R_(1))-1/(R_(2)))`

Explanation

`(mu_(0)I)/4(1/(R_(1))-1/(R_(2)))`
At C magnetic field due to wires PQ and PQ and RS will be zero. Due to wire QR,
`B_(1)=1/2 ((mu_(0)I)/(2R_(1)))=(mu_(0)I)/(4R_(1))` (perpendicular to paper outwards) , And due to wire SP,
`B_(2)=1/2((mu_(0)I)/(2R_(2)))=(mu_(0)I)/(4R_(2))` (perpendicular to paper inwards), `:.` Net Magnetic field would be
`B=(mu_(0)I)/4(1/(R_(1))-1/(R_(2)))` (Perpendicular to paper outwards)