Wires 1 and 2 carrying currents i_1 and i_2 r | vedclass

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Question

(D) The magnetic field $B$ produced by the long straight wire $1$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i_1}{2\pi r}$.The force

Vedclass · Accepted Answer

$\frac{\mu_0}{2\pi r} i_1 i_2 \, dl$

Vedclass · Answer

(D) The magnetic field $B$ produced by the long straight wire $1$ at a perpendicular distance $r$ is given by $B = \frac{\mu_0 i_1}{2\pi r}$.The force $dF$ on a current element $i_2 \, dl$ placed in a magnetic field $B$ is given by $dF = i_2 (dl 	imes B) = i_2 \, dl \, B \sin \alpha$,where $\alpha$ is the angle between the current element $dl$ and the magnetic field $B$.The magnetic field $B$ is perpendicular to the plane containing the wires (directed inwards). The current element $dl$ lies in the plane of the wires. Therefore,the angle between $dl$ and $B$ is $90^{\circ}$.Thus,$dF = i_2 \, dl \left( \frac{\mu_0 i_1}{2\pi r} ight) \sin 90^{\circ} = \frac{\mu_0 i_1 i_2}{2\pi r} dl$.

Wires $1$ and $2$ carrying currents $i_1$ and $i_2$ respectively are inclined at an angle $\theta$ to each other. What is the force on a small element $dl$ of wire $2$ at a distance of $r$ from wire $1$ (as shown in figure) due to the magnetic field of wire $1$?

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