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.

(a)

Consider the following diagram,

As the given frame is at rest, electric field on$q_B$ due to charge$q_A$ is,

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

Here$d$ is the distance between the charge,$q_A$ and$q_B$ are the charges.

Charge$q_A$ is at rest, therefore the magnetic field due to charge $q_A$at the position charge $q_B$is zero.

Now the force acting on charge $q_B$is given by,

$F=q_{B}E$

Substitute$\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q_A}{d^2}\stackrel{âˆ\S }{y}$ for $E$in the above equation

$F=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q_A{q}_B}{d^2}\stackrel{âˆ\S }{y}$

Hence, the electromagnetic force on$q_B$ is$F=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q_A{q}_B}{d^2}\stackrel{âˆ\S }{y}$ .

(b)

(i)

Consider the frame$\stackrel{→}{S}$as a accelerated frame of reference, moving with velocity$v$.

Since, the particle is at rest in$\stackrel{→}{S}$frame of reference.

The expression for the force on $q_B$when$q_A$passes the $y$axis is,

$F=q_{B}E$

Here $F$is the force,$q_B$is the charge and$E$is the electric field.

Now using the special theory of relativity, electric field lines will contract by a factor if the point charge is moving.

Let us assume that the point charge is moving with speed then electric field lines will contract by$\sqrt{1−\frac{v^2}{c^2}}$.

The expression for the force on charge $q_B$due to contracted field lines,

$F=q_B{E}^{\text{'}}$

Substitute $\frac{\mathrm{γ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q_A}{d^2}\stackrel{âˆ\S }{y}$for$E^{\text{'}}$in the above equation,

$F=q_B\frac{\mathrm{γ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q_A}{d^2}\stackrel{âˆ\S }{y}$

Here $\mathrm{γ}=\frac{1}{\sqrt{1−\frac{v^2}{c^2}}}$

Therefore, the force on$q_B$when $q_A$passes the $\stackrel{→}{y}$axis by transforming the force is $\stackrel{→}{F}=\frac{\mathrm{γ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q_A{q}_B}{d^2}\stackrel{âˆ\S }{y}$.

(ii)

The expression for the electric field due to charge $q_A$in frame is,$\stackrel{→}{S}$

$E=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q(1−\frac{v^2}{c^2})}{{[1−\left(\frac{v^2}{c^2}\right){\sin }^2\mathrm{θ}]}^{\frac{3}{2}}{d}^2}\stackrel{âˆ\S }{r}$

As the particle is moving parallel to$x$ axis,

Substitute$90°$ for$\mathrm{θ}$ in the above equation,

$\begin{array}{c}E=\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q(1−\frac{v^2}{c^2})}{{[1−\left(\frac{v^2}{c^2}\right){\sin }^{2}90°]}^{\frac{3}{2}}{d}^2}\stackrel{âˆ\S }{r}\\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q(1−\frac{v^2}{c^2})}{{[1−\left(\frac{v^2}{c^2}\right)]}^{\frac{3}{2}}{d}^2}\stackrel{âˆ\S }{r}\\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q}{{[1−\left(\frac{v^2}{c^2}\right)]}^{\frac{1}{2}}{d}^2}\stackrel{âˆ\S }{r}\end{array}$

The expression for the force on $q_B$when $q_A$passes the$\stackrel{→}{y}$ is,

$\stackrel{→}{F}=\frac{\mathrm{γ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q_A{q}_B}{d^2}\stackrel{âˆ\S }{y}$

Here $\mathrm{γ}=\frac{1}{\sqrt{1−\frac{v^2}{c^2}}}$

As the frame$\stackrel{→}{S}$ is at rest, thus $v=0$so, $\stackrel{→}{B}=0$,

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

$\stackrel{→}{F}=q_B[E+(\stackrel{→}{v}×B)]$

Substitute$\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q}{{[1−\left(\frac{v^2}{c^2}\right)]}^{\frac{1}{2}}{d}^2}\stackrel{âˆ\S }{y}$ for$E$ and $E$for$B$ in the above equation.

$\begin{array}{c}\stackrel{→}{F}=q_B[\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q_A}{{[1−\left(\frac{v^2}{c^2}\right)]}^{\frac{1}{2}}{d}^2}\stackrel{âˆ\S }{y}+0]\\ =\frac{1}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q_B{q}_A}{{[1−\left(\frac{v^2}{c^2}\right)]}^{\frac{1}{2}}{d}^2}\stackrel{âˆ\S }{y}\\ =\frac{\mathrm{γ}{q}_B{q}_A}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{1}{d^2}\stackrel{âˆ\S }{y}\end{array}$

Therefore the force on $q_B$when $q_A$passes the $\stackrel{→}{y}$axis by computing the fields in $\stackrel{→}{S}$and using the Lorentz law is $\stackrel{→}{F}=\frac{\mathrm{γ}}{4\mathrm{Ï€}{\mathrm{Î\mu }}_o}\frac{q_A{q}_B}{d^2}\stackrel{âˆ\S }{y}$.