What is the magnetic field at point P in the | vedclass

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Question

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

Vedclass · Accepted Answer

$\frac{{{\mu _0}i}}{{\pi r}}$

Vedclass · Answer

(A) The magnetic field due to a finite straight 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,the perpendicular distance from point $P$ to each wire is $d = r \sin {45^o} = \frac{r}{{\sqrt 2 }}$.For each wire,the angles subtended at point $P$ are ${	heta _1} = {0^o}$ (at the corner $O$) and ${	heta _2} = {45^o}$.The magnetic field due to one wire is $B_1 = \frac{{{\mu _0}i}}{{4\pi (r/\sqrt 2 )}}(\sin {0^o} + \sin {45^o}) = \frac{{{\mu _0}i \sqrt 2 }}{{4\pi r}} 	imes \frac{1}{{\sqrt 2 }} = \frac{{{\mu _0}i}}{{4\pi r}}$.Since both wires carry current in directions such that their magnetic fields at $P$ point in the same direction (into the page),the net magnetic field is $B_{net} = B_1 + B_2 = 2 	imes \frac{{{\mu _0}i}}{{4\pi r}} = \frac{{{\mu _0}i}}{{2\pi r}}$.However,re-evaluating the geometry based on the provided solution image,the angles are ${	heta _1} = {45^o}$ and ${	heta _2} = {45^o}$ relative to the perpendicular. The formula $B = \frac{{{\mu _0}i}}{{4\pi d}}(\sin {	heta _1} + \sin {	heta _2})$ with $d = r/\sqrt{2}$ gives $B = \frac{{{\mu _0}i}}{{4\pi (r/\sqrt{2})}}(\sin 45^o + \sin 45^o) = \frac{{{\mu _0}i \sqrt{2}}}{4\pi r} (2 	imes \frac{1}{\sqrt{2}}) = \frac{{{\mu _0}i}}{{2\pi r}}$.Given the options,the intended answer is $B = \frac{{{\mu _0}i}}{{2\pi r}}$ which matches option $A$.

What is the magnetic field at point $P$ in the given figure?

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