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Question

(B) The magnetic field due to a finite wire at a perpendicular distance $d$ is given by $B = \frac{{\mu _0 i}}{{4\pi d}}(\sin 	heta _1 + \sin 	heta

Vedclass · Accepted Answer

$\frac{{\mu _0 i}}{{8\sqrt 3 \pi l}}$

Vedclass · Answer

(B) The magnetic field due to a finite wire at a perpendicular distance $d$ is given by $B = \frac{{\mu _0 i}}{{4\pi d}}(\sin 	heta _1 + \sin 	heta _2)$.From the geometry of the figure:$1$. The perpendicular distance from the wire to point $P$ is $d = BP = \sqrt{3}l$.$2$. The angle at end $B$ is $	heta _1 = 0^\circ$ (since $P$ lies on the line perpendicular to the wire at $B$).$3$. The angle at end $A$ is $	heta _2$. In $	riangle ABP$,$	an 	heta _2 = \frac{AB}{BP} = \frac{l}{\sqrt{3}l} = \frac{1}{\sqrt{3}}$. Thus,$	heta _2 = 30^\circ$.Substituting these values into the formula:$B_P = \frac{{\mu _0 i}}{{4\pi (\sqrt{3}l)}}(\sin 0^\circ + \sin 30^\circ)$$B_P = \frac{{\mu _0 i}}{{4\sqrt{3}\pi l}}(0 + \frac{1}{2})$$B_P = \frac{{\mu _0 i}}{{8\sqrt{3}\pi l}}$

The magnetic field at point $P$ is:

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