A long wire bent as shown in fig. carries current I. If the radius of the semicircular portion is a, the magnetic field at centre C is

Options

A. `(mu_0I)/(4a)`
B. `(mu_0I)/(4pia)sqrt(pi^2+4)`
C. `(mu_0I)/(4a)+(mu_0I)/(4pia)`
D. `(mu_0I)/(4pia)sqrt((pi^2-4))`

Answer

B

Explanation

(b) `B_1=(mu_0)/(4pi)xx(2piI)/axx1/2 ("due to semicircular part")`
`B_2=(mu_0)/(4pi)xx(2I)/a ("due to parallel parts of currents")`
These two fields are at right angles to each other.
Hence, resultant field
`B=sqrt(B_1^2+B_2^2)=(mu_0I)/(4pia)sqrt(pi^2+4)`