Question

Two very long straight parallel wires having current $I$ and $2I$ as shown in the figure. A point charge $q$ is at a point equidistant from the two wires in the plane of the wires. Its instantaneous velocity $\overline{v}$ is parallel to currents in the plane. The magnitude of force due to the magnetic field acting on the charge at this instant is:

• A. $\frac{qv{\mu }_{0}l}{2\pi d}$
• B. $\frac{2qv{\mu }_{0}l}{3\pi d}$
• C. $\frac{6qv{\mu }_{0}l}{2\pi d}$
• D. $\frac{qv{\mu }_{0}l}{\pi d}$

Answer 1

The correct option is D $\frac{qv{\mu }_{0}l}{\pi d}$

Magnetic force on $q$

$\overrightarrow{-B}net\times \overrightarrow{V}\times q$

$=\overrightarrow{V}\times \overrightarrow{B}net\times q$

$\overrightarrow{B}net=\overrightarrow{B_1}+\overrightarrow{B_2}$

$\overrightarrow{B_1}=\frac{roI}{2n\left(\frac{d}{2}\right)}=\frac{roI}{nd}(\times )$

$\overrightarrow{B2}=\frac{2roI}{2n\left(\frac{d}{2}\right)}=\frac{2roI}{nd}\odot$

$\overrightarrow{B}net=\overrightarrow{B_1+}\overrightarrow{B_2}=\frac{roI}{nd}\odot$

So,

force $=\frac{roI}{nd}\left(V\right)q[\therefore \overrightarrow{V}and\overrightarrow{B}netare\perp ]$

force is acting foward wire $-2.$

option $-D$ is correct.