The straight wire A B 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

Correct Answer is : µ0I4πd(sinθ1−sinθ2)

\frac{\mu_0 I}{4 \pi d} ( \sin \theta_1- \sin \theta_2)

\frac{\mu_0 I}{4 \pi d} ( \sin \theta_1+ \sin \theta_2)

\frac{\mu_0 I}{4 \pi d} (\cos \theta_1-\cos \theta_2)

\frac{\mu_0 I}{4 \pi d} (\cos \theta_1+\cos \theta_2)

BITSAT 2024 Shift 1 Paper

Question : 38 of 130

Marks: +1, -0

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

‌
µ0I
4πd
(sin‌θ1−sin‌θ2)
‌
µ0I
4πd
(sin‌θ1+sin‌θ2)
‌
µ0I
4πd
(cos‌θ1−cos‌θ2)
‌
µ0I
4πd
(cos‌θ1+cos‌θ2)

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