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(D) The magnetic field $B$ due to a finite straight wire at a perpendicular distance $R$ from the wire is given by $B = \frac{\mu_0 i}{4\pi R}(\sin

Vedclass · Accepted Answer

$Zero$

Vedclass · Answer

(D) The magnetic field $B$ due to a finite straight wire at a perpendicular distance $R$ from the wire is given by $B = \frac{\mu_0 i}{4\pi R}(\sin 	heta_1 + \sin 	heta_2).$Here, the point is on the axis of the conductor. The axis of a straight conductor is the line passing through the conductor itself.For any point lying on the axis of a straight current-carrying conductor, the angle between the position vector of the point and the current element is $0^\circ$ or $180^\circ$.According to the Biot-Savart Law, $dB = \frac{\mu_0 i}{4\pi} \frac{dl \sin 	heta}{r^2}$.Since $	heta = 0^\circ$ or $180^\circ$, $\sin 	heta = 0$, which implies $dB = 0$.Therefore, the magnetic field at any point on the axis of a straight current-carrying conductor is $Zero$.

$A$ current $i$ is flowing in a straight conductor of length $L.$ The magnetic induction at a point on its axis at a distance $\frac{L}{4}$ from its centre will be

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