Q. The straight wire AB carries a current I. The ends of the wire subtend angles θ1 and θ2 at the point P as shown in figure. The magnetic field at the point P is
(a) $ \displaystyle \frac{\mu_0 I}{4\pi a} (sin\theta_1 – sin\theta_2)$
(b) $ \displaystyle \frac{\mu_0 I}{4\pi a} (sin\theta_1 + sin\theta_2)$
(c) $\displaystyle \frac{\mu_0 I}{4\pi a} (coos\theta_1 – cos\theta_2)$
(d) $\displaystyle \frac{\mu_0 I}{4\pi a} (coos\theta_1 + cos\theta_2)$
Ans: (a)