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(A) The magnetic field $B$ at a perpendicular distance $r$ from a finite wire carrying current $I$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin 	he

Vedclass · Accepted Answer

$\frac{\mu_0 I}{2 \sqrt{3} \pi x}$

Vedclass · Answer

(A) The magnetic field $B$ at a perpendicular distance $r$ from a finite wire carrying current $I$ is given by $B = \frac{\mu_0 I}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$.In the given figure,the perpendicular distance $r$ from point $P$ to the wire is $r = x \cos 30^{\circ} = x \frac{\sqrt{3}}{2}$.The angles subtended by the ends of the wire at point $P$ are $	heta_1 = 30^{\circ}$ and $	heta_2 = 30^{\circ}$.Substituting these values into the formula:$B = \frac{\mu_0 I}{4 \pi (x \frac{\sqrt{3}}{2})} (\sin 30^{\circ} + \sin 30^{\circ})$$B = \frac{\mu_0 I}{2 \sqrt{3} \pi x} (\frac{1}{2} + \frac{1}{2})$$B = \frac{\mu_0 I}{2 \sqrt{3} \pi x}$

$A$ straight wire of finite length carrying current $I$ subtends an angle of $60^{\circ}$ at point $P$ as shown. The magnetic field at $P$ is

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