91Ó°ÊÓ

The expression for the force from the Coulomb’s law is given by;

$F=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q_1{q}_2}{r^2}$

Here$F$ is the magnitude of force between two charge,$q_1$ ,$q_2$ are the magnitude of the charges and $r$is the distance between the two charge.

The expression for the magnetic field is given by,

$\stackrel{→}{B}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\left(\frac{Q(\stackrel{→}{v}×\stackrel{→}{r})}{r^2}\right)$

Here$\stackrel{→}{B}$ is the magnetic field, $\stackrel{→}{v}$is the velocity of the charge,$Q$ is the magnitude of the charge,$\stackrel{→}{r}$ is the position vector of the charge and${\mathrm{μ}}_o$ is the permeability of the free space.

For the given system the position of the charge $+q$is $−\frac{d}{2}\stackrel{âˆ\S }{y}$and position of $−q$is $\frac{d}{2}\stackrel{âˆ\S }{y}$.

Electric field

The general expression for the electric field is,

$\stackrel{→}{E}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{Q}{r^2}\stackrel{âˆ\S }{r}$

System A;

Substitute $−q$for $Q$,$\stackrel{âˆ\S }{y}$for $\stackrel{âˆ\S }{r}$and $d$for$r$in the above equation.

$\stackrel{→}{E}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q}{d^2}\stackrel{âˆ\S }{y}$

The motion of the charges will not affect the electric field at position of $+q$due to $−q$.

System B;

$\stackrel{→}{E}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q}{d^2}\stackrel{âˆ\S }{y}$

System C;

$\stackrel{→}{E}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q}{d^2}\stackrel{âˆ\S }{y}$

Thus, for systems, electric field at $+q$due to is $\stackrel{→}{E}=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q}{d^2}\stackrel{âˆ\S }{y}$.

Magnetic field

The general expression for the magnetic field is given by,

$\stackrel{→}{B}=\frac{{\mathrm{μ}}_o}{4\mathrm{π}}\left(\frac{Q(\stackrel{→}{v}×\stackrel{→}{r})}{r^2}\right)$

System A;

Substitute $−q$for $Q$, $\stackrel{âˆ\S }{y}$for $\stackrel{âˆ\S }{r}$,$v(−\stackrel{âˆ\S }{x})$ for $\stackrel{→}{v}$and $d$for $r$in the above equation.

role="math" localid="1658131922476" $\begin{array}{c}\stackrel{→}{B}=\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\left(\frac{−q(v(−\stackrel{âˆ\S }{x})×\stackrel{âˆ\S }{y})}{d^2}\right)\\ =\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\left(\frac{−qv(−\stackrel{âˆ\S }{z})}{d^2}\right)\\ =\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\left(\frac{qv}{d^2}\right)\stackrel{âˆ\S }{z}\end{array}$

System B;

Now if the charge $−q$is moving and$+q$is at rest, the magnetic field at the position of charge$+q$is produced due to the motion of charge $−q$.

The magnetic field produced at the position of will be same as calculated in system A.

$\stackrel{→}{B}=\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\left(\frac{qv}{d^2}\right)\stackrel{âˆ\S }{z}$

System C;

As the charge$−q$is at rest and$+q$is at moving. The magnetic field at position of charge $+q$is zero as magnetic field is produced due to the motion of charge $−q$

$\stackrel{→}{B}=0$

Thus magnetic field at system A and B is $\stackrel{→}{B}=\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\left(\frac{qv}{d^2}\right)\stackrel{âˆ\S }{z}$and at system C is $\stackrel{→}{B}=0$.

Force

The expression for the force using Lorentz law is given by,

$\stackrel{→}{F}=Q(\stackrel{→}{E}+\stackrel{→}{v}×\stackrel{→}{B})$...... (1)

System A;

Substitute$+q$for$Q$, $\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q}{d^2}\stackrel{âˆ\S }{y}$for$\stackrel{→}{E}$,$−v\left(\stackrel{âˆ\S }{x}\right)$for$\stackrel{→}{v}$ and $\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\left(\frac{qv}{d^2}\right)\stackrel{âˆ\S }{z}$for $\stackrel{→}{B}$in the equation (1).

$\begin{array}{c}\stackrel{→}{F}=q\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q}{d^2}\stackrel{âˆ\S }{y}\right)+−v\left(\stackrel{âˆ\S }{x}\right)×\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\left(\frac{qv}{d^2}\right)\stackrel{âˆ\S }{z}\\ =\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q^2}{d^2}\stackrel{âˆ\S }{y}\right)+\left(\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\left(\frac{q^2{v}^2}{d^2}\stackrel{âˆ\S }{y}\right)\right)\end{array}$

System B;

The velocity of the charge $+q$is zero.

Substitute $+q$for $Q$,$\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q}{d^2}\stackrel{âˆ\S }{y}$for $\stackrel{→}{E}$,$0$for $\stackrel{→}{v}$and $\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\left(\frac{qv}{d^2}\right)\stackrel{âˆ\S }{z}$for$\stackrel{→}{B}$in the equation (1).

role="math" localid="1658133092641" $\begin{array}{c}\stackrel{→}{F}=q\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q}{d^2}\stackrel{âˆ\S }{y}\right)+−0×\frac{{\mathrm{μ}}_o}{4\mathrm{Ï€}}\left(\frac{qv}{d^2}\right)\stackrel{âˆ\S }{z}\\ =\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q^2}{d^2}\stackrel{âˆ\S }{y}\right)\end{array}$

System C;

The velocity of the charge$−q$is zero.

Substitute $+q$for $Q$, $\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q}{d^2}\stackrel{âˆ\S }{y}$for $\stackrel{→}{E}$, $v(−\stackrel{âˆ\S }{x})$for $\stackrel{→}{v}$ and for $\stackrel{→}{B}$in the equation (1).

.$\begin{array}{c}\stackrel{→}{F}=q\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q}{d^2}\stackrel{âˆ\S }{y}\right)+v(−\stackrel{âˆ\S }{x})×0\\ =\left(\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{−q^2}{d^2}\stackrel{âˆ\S }{y}\right)\end{array}$

Therefore,