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_(0)I)/(2R) o.`
B. `(mu_(0)I)/(2R)(1+1/(pisqrt(2))) o.`
C. `(mu_(0)I)/(2R)(1-1/(pisqrt(2))) ox`
D. `(mu_(0)I)/(R)(1-1/(pisqrt(2))) ox`

Answer

B

Explanation

Magnetic field due to `'AB'`
`vec(B)_(1)=(mu_(0)I)/(4piR)[sin(pi/2)-sin(pi/4)] ox`
`=(mu_(0)I)/(4piR)[1-1/(sqrt(2))]ox`
Magnetic field due to circualr loop
`vec(B)_(2)=(mu_(0)I)/(2R) o.`
Magnetic field due to straight wire `BC`
`vec(B)_(3)=(mu_(0)I)/(4piR)[sin(pi/2)+sin(pi/4)] o.`
`=(mu_(0)I)/(4piR)[1+1/(sqrt(2))]o.`
`vec(B)=vec(B)_(1)+vec(B)_(2)+vec(B)_(3)`
`=((mu_(0)I)/(2R)+(2mu_(0)I)/(2piR) 1/(2sqrt(2))) o.=(mu_(0)I)/(2R)[1+1/(sqrt(2)pi)]`