Two parallel wires situated at a distance 2a | vedclass

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(A) The magnetic field due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 i}{2\pi r}$.Since the currents are in opposite direc

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$\frac{\mu_0 ia}{\pi r^2}$

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(A) The magnetic field due to a long straight wire at a distance $r$ is given by $B = \frac{\mu_0 i}{2\pi r}$.Since the currents are in opposite directions,the magnetic field vectors $B_1$ and $B_2$ at point $P$ will have the same magnitude $B = \frac{\mu_0 i}{2\pi r}$.From the geometry of the figure,the vertical components of the magnetic fields cancel each other,while the horizontal components add up.The angle between the horizontal axis and the magnetic field vector is $	heta$,where $\sin 	heta = \frac{a}{r}$.The resultant magnetic field $B_{net} = B_1 \cos 	heta + B_2 \cos 	heta = 2B \cos 	heta$.Since $\cos 	heta = \sqrt{1 - \sin^2 	heta} = \sqrt{1 - (a/r)^2} = \frac{\sqrt{r^2 - a^2}}{r}$.However,looking at the geometry,the horizontal component is $B \sin 	heta$ (based on the angle $	heta$ shown with the vertical). Let's re-evaluate: The angle $	heta$ is between the vertical and the field vector. The horizontal component is $B \sin 	heta$.$B_{net} = 2B \sin 	heta = 2 \left( \frac{\mu_0 i}{2\pi r} ight) \left( \frac{a}{r} ight) = \frac{\mu_0 ia}{\pi r^2}$.

Two parallel wires situated at a distance $2a$ are carrying equal currents $i$ in opposite directions as shown in the figure. The value of the magnetic field at a point $P$ situated at equal distances $r$ from both the wires will be:

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