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Question

(A) The magnetic field due to a semi-infinite wire at a perpendicular distance $r$ is given by $B = \frac{\mu_{0} I}{4 \pi r} (\sin 	heta_1 + \sin

Vedclass · Accepted Answer

$\frac{\mu_{0} I}{4 \pi x y}\left[\sqrt{x^{2}+y^{2}}+(x+y)\right]$

Vedclass · Answer

(A) The magnetic field due to a semi-infinite wire at a perpendicular distance $r$ is given by $B = \frac{\mu_{0} I}{4 \pi r} (\sin 	heta_1 + \sin 	heta_2)$.For wire $(1)$ (along the $x$-axis), the distance from $P$ is $y$. One end is at the origin $(	heta_1 = 90^{\circ})$ and the other is at infinity $(	heta_2 = 90^{\circ})$, but since it is a semi-infinite wire starting from the origin, the formula becomes $B_1 = \frac{\mu_{0} I}{4 \pi y} (1 + \sin 	heta_1)$, where $\sin 	heta_1 = \frac{x}{\sqrt{x^2+y^2}}$.Thus, $B_1 = \frac{\mu_{0} I}{4 \pi y} \left(1 + \frac{x}{\sqrt{x^2+y^2}}ight)$.Similarly, for wire $(2)$ (along the $y$-axis), the distance from $P$ is $x$, so $B_2 = \frac{\mu_{0} I}{4 \pi x} \left(1 + \frac{y}{\sqrt{x^2+y^2}}ight)$.Both fields are directed into the page at point $P$. Adding them:$B = B_1 + B_2 = \frac{\mu_{0} I}{4 \pi} \left[ \frac{1}{y} + \frac{x}{y\sqrt{x^2+y^2}} + \frac{1}{x} + \frac{y}{x\sqrt{x^2+y^2}} ight]$$B = \frac{\mu_{0} I}{4 \pi} \left[ \frac{x+y}{xy} + \frac{x^2+y^2}{xy\sqrt{x^2+y^2}} ight]$$B = \frac{\mu_{0} I}{4 \pi xy} \left[ (x+y) + \sqrt{x^2+y^2} ight]$.

There are two infinitely long straight current-carrying conductors held at right angles to each other such that their common ends meet at the origin, as shown in the figure. The ratio of current in both conductors is $1:1$. The magnetic field at point $P(x, y)$ is:

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