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

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Question

(B) The magnetic field $B$ produced by the long straight wire $1$ carrying current $i_1$ at a distance $r$ is given by the formula $B = \frac{\mu_0 i_

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The magnetic field $B$ produced by the long straight wire $1$ carrying current $i_1$ at a distance $r$ is given by the formula $B = \frac{\mu_0 i_1}{2\pi r}$.This magnetic field is directed perpendicular to the plane containing the wires.The force $dF$ on a small current element $i_2 \vec{dl}$ placed in a magnetic field $\vec{B}$ is given by $d\vec{F} = i_2 (\vec{dl} 	imes \vec{B})$.The magnitude of this force is $dF = i_2 dl B \sin \alpha$,where $\alpha$ is the angle between the current element $dl$ and the magnetic field $B$.Since the magnetic field $B$ is perpendicular to the plane of the wires,it is perpendicular to the current element $dl$ as well. Therefore,$\alpha = 90^\circ$ and $\sin 90^\circ = 1$.Thus,the magnitude of the force is $dF = i_2 dl B = i_2 dl \left( \frac{\mu_0 i_1}{2\pi r} ight) = \frac{\mu_0 i_1 i_2}{2\pi r} dl$.

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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