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

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(C) The magnetic field $B$ produced by wire $1$ at a distance $r$ is given by $B = \frac{\mu_0 i_1}{2\pi r}$.This magnetic field is directed perpendic

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The magnetic field $B$ produced by wire $1$ at a distance $r$ is given by $B = \frac{\mu_0 i_1}{2\pi r}$.This magnetic field is directed perpendicular to the plane containing the wires.The force $dF$ on a current element $i_2 dl$ in a magnetic field $B$ is given by $dF = i_2 (dl 	imes B)$.The magnitude of the force is $dF = i_2 B dl \sin \phi$,where $\phi$ 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. Thus,$\phi = 90^\circ$ and $\sin \phi = 1$.However,the force between two current-carrying wires is specifically due to the component of the current element that is parallel to the other wire.The component of $dl$ parallel to wire $1$ is $dl \cos 	heta$.The force on this component is $dF = B \cdot i_2 \cdot (dl \cos 	heta) = \left( \frac{\mu_0 i_1}{2\pi r} ight) i_2 dl \cos 	heta = \frac{\mu_0 i_1 i_2 dl \cos 	heta}{2\pi r}$.

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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