The straight wire `AB` carries a current `I`. The ends of the wire subtend angles `theta_1` and `theta_2` at the point `P` as shown in figure. The magnetic field at the point `P` is

Options

A. `(mu_(0)I)/(4pia)(sin theta_(1) -sin theta_(2))`
B. `(mu_(0)I)/(4 pi a)(sin theta_(1)+sin theta_(2))`
C. `(mu_(0)I)/(4pia)(cos theta_(1)-cos theta_(2))`
D. `(mu_(I))/(4pia)(cos theta_(1)+cos theta_(2))`

Answer

A

Explanation

`B=(mu_(0)I)/(4pia) [ sin theta_(1)+sin (-theta_(2))]`