A wire of uniform cross section is used to make the given system. The radius of semicircle BCD is R and that of AFE is 2R while current I enters at point A.

Find the magnetic field intensity at point O in the figure if resistance per unit length of wire is `lambda`.

Explanation

Let `R_(1)` be the resistance of ABCDE and `R_(2)` be that of AFE.
` R_(1) = lambdaR + lambda piR + lambdaR = lambdaR (2 + pi)`
` R_(2) = lamda[pi (2 R)] = 2 (lambda pi R)`
Both sides are connected in parallel and hence for the same potential difference, the current will be divided between them as follows:
`i_(1) = R_(2)/(R_(1) + R_(2)) i = (2 lambdapiR)/(lambdaR(2 + pi)+2lambdapiR) i = (2pi)/(2 +3 pi) i`
`i_(2) = R_(1)/(R_(1) + R_(2)) i = (lambdaR(2 + pi))/(lambdaR(2 + pi) + 2 lambdapiR) i = (2 + pi)/(2 + 3 pi) i`
Straight portions will not create any maxmetic field at the centre. Magnetic field due to the semicircle with bigger radius is outward and that due to the semicircle with smaller radius is inward. Moreover, we have to use magnetic field due to semicircle as `B = (mu_(0)i)/(4 r) ` .
`B_(1) = mu_(0)/(4R) ((2pi)/(2 + 3pi) i)`
` B_(2) = mu_(0)/(4(2R)) ((2 + pi)/(3 + 3pi) i) `
The net magnetic field at the centre can be written as follows:
`B = mu_(0)/(4(R)) ((2pi)/(2 + 3pi)i) - mu_(0)/(4(2R)) ((2 + pi)/(2 + 3pi) i ) `
`B = mu_(0)/(8R) [(4pi)/(2 + 3pi) - (2 + pi)/(2 + 3pi)]i = mu_(0)/(8R) [(4pi - 2 - pi)/(2 + 3pi)]i`
`B = mu_(0)/(8R) [(3 pi - 2)/(3pi + 2)] i `