The magnetic field at the centre of the circular loop as shown in Fig. when a single wire is bent to form a circular loop and also extends to form straight sections is

Options

A. `(mu_0I)/(2R)odot`
B. `(mu_0I)/(2R) (1+1/(pisqrt2))odot`
C. `(mu_0I)/(2R) (1-1/(pisqrt2))ox`
D. `(mu_0I)/R (1-1/(pisqrt2))ox`

Answer

B

Explanation

(b) Magnetic field due to 'AB'
`vecB_1=(mu_0I)/(4piR)[sinpi/2-sinpi/4]ox=(mu_0I)/(4piR)[1-1/(sqrt2)]ox`
Magnetic field due to circular loop,
`vecB_2=(mu_0I)/(2R)odot`
Magnetic field due to straight wire BC,
`vecB_3=(mu_0I)/(4piR)[sinpi/2-sinpi/4]odot=(mu_0I)/(4piR)[1+1/(sqrt2)]odot`
`vecB=vecB_1+vecB_2+vecB_3`
`=((mu_0I)/(2R)+(2mu_0I)/(4piR) 1/(sqrt2))odot=(mu_0I)/(2R)[1+1/(sqrtpi)]odot`