For the situation shown in the figure, flux through the square loop is:

Options

A. `((mu_(0)ia)/(2pi))ln(a/(2a-b))`
B. `((mu_(0)ib)/(2pi))ln(a/(2b-a))`
C. `((mu_(0)ib)/(2pi))ln(a/(b-a))`
D. `((mu_(0)ia)/(2pi))ln((2a)/(b-a))`

Answer

C

Explanation

`intdphi=int(mu_(0)I)/(2pix)(bdx)`
`phi=(mu_(0)Ib)/(2pi)underset((b+a))overset(a)int(dx)/x`
`phi=(mu_(0)Ib)/(2pi)ln(a/(b-a))`