The wire shown in figure carries a current of `32 A`. If `r = 3.14 cm`, the magnetic field at point `P` will be

Options

A. `1.6 xx 10^(-4) T`
B. `3.2 xx 10^(-4) T`
C. `4.8 xx 10^(-4) T`
D. `6.4 xx 10^(-4) T`

Answer

C

Explanation

(c ) Magnetic field due to a complete circular wire loop carrying a currrent `I` at the centre is `(mu_(0) I)/(2 r)`.
Here, the straight portions of the wire do not contribute because the point `P` is along them. The field at `P` is due to `(3)/(4)th ` fo the loop of radius `r`. Thus
`B = (3)/(4) ((mu_(0) I)/(2 r))`
`= (3)/(4) xx (4 pi xx 10^(-7) xx 32)/(2 xx 3.14 xx 10^(-7)) = 4.8 xx 10^(-4) T`