The wire loop formed by joining two semicircular sections of radii `R_(1)` and `R_(2)` and centre C, carries a current I as shown. The magnetic field at C has a magnitude

Options

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

Answer

B

Explanation

Magnetic field at the centre C of a semicircular conductor is given by
`B=sumdB`
`=sum(mu_(0))/(4pi)(Idlsin90^(@))/(r^(2))`
`=(mu_(0))/(4pi)(I)/(r^(2))sumdl=(mu_(0))/(4pi)*(I)/(r^(2))(pir)=(mu_(0)I)/(4r)`
Here, magnetic field at C due to the current in the circular loop of smaller radius, `vecB_(1)=(mu_(0)I)/(4R_(1))` upwards perpendicular to the plane of the paper.
Magnetic to the plane of the paper.
Magnetic field at C due to the current (I) in the circular loop of bigger radius, `vecB_(2)=(mu_(0)I)/(4R_(2))` downwards perpendicular to the plane of the paper.
Also magnetic field due to straight portions is zero.
Hence, net magnetic field at C
`=vecB_(1)+vecB_(2)=|vecB_(1)|-|vecB_(2)|`
`=(mu_(0)I)/(4R_(1))-(mu_(0)I)/(4R_(2))=(mu_(0)I)/(4)[(1)/(R_(1))-(1)/(R_(2))]` upwards.