91影视

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

$F=\frac{1}{4蟺{蔚}_o}\frac{q_1{q}_2}{r^2}$

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,

$\stackrel{鈫�}{B}=\frac{{渭}_o}{4蟺}\left(\frac{Q(\stackrel{鈫�}{v}脳\stackrel{鈫�}{r})}{r^2}\right)$

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

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

Electric field

The general expression for the electric field is,

$\stackrel{鈫�}{E}=\frac{1}{4蟺{蔚}_o}\frac{Q}{r^2}\stackrel{鈭�}{r}$

System A;

Substitute $鈭�q$for $Q$,$\stackrel{鈭�}{y}$for $\stackrel{鈭�}{r}$and $d$for$r$in the above equation.

$\stackrel{鈫�}{E}=\frac{1}{4蟺{蔚}_o}\frac{鈭�q}{d^2}\stackrel{鈭�}{y}$

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

System B;

$\stackrel{鈫�}{E}=\frac{1}{4蟺{蔚}_o}\frac{鈭�q}{d^2}\stackrel{鈭�}{y}$

System C;

$\stackrel{鈫�}{E}=\frac{1}{4蟺{蔚}_o}\frac{鈭�q}{d^2}\stackrel{鈭�}{y}$

Thus, for systems, electric field at $+q$due to is $\stackrel{鈫�}{E}=\frac{1}{4蟺{蔚}_o}\frac{鈭�q}{d^2}\stackrel{鈭�}{y}$.

Magnetic field

The general expression for the magnetic field is given by,

$\stackrel{鈫�}{B}=\frac{{渭}_o}{4蟺}\left(\frac{Q(\stackrel{鈫�}{v}脳\stackrel{鈫�}{r})}{r^2}\right)$

System A;

Substitute $鈭�q$for $Q$, $\stackrel{鈭�}{y}$for $\stackrel{鈭�}{r}$,$v(鈭�\stackrel{鈭�}{x})$ for $\stackrel{鈫�}{v}$and $d$for $r$in the above equation.

role="math" localid="1658131922476" $\begin{array}{c}\stackrel{鈫�}{B}=\frac{{渭}_o}{4蟺}\left(\frac{鈭�q(v(鈭�\stackrel{鈭�}{x})脳\stackrel{鈭�}{y})}{d^2}\right)\\ =\frac{{渭}_o}{4蟺}\left(\frac{鈭�qv(鈭�\stackrel{鈭�}{z})}{d^2}\right)\\ =\frac{{渭}_o}{4蟺}\left(\frac{qv}{d^2}\right)\stackrel{鈭�}{z}\end{array}$

System B;

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

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

$\stackrel{鈫�}{B}=\frac{{渭}_o}{4蟺}\left(\frac{qv}{d^2}\right)\stackrel{鈭�}{z}$

System C;

As the charge$鈭�q$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 $鈭�q$

$\stackrel{鈫�}{B}=0$

Thus magnetic field at system A and B is $\stackrel{鈫�}{B}=\frac{{渭}_o}{4蟺}\left(\frac{qv}{d^2}\right)\stackrel{鈭�}{z}$and at system C is $\stackrel{鈫�}{B}=0$.

Force

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

$\stackrel{鈫�}{F}=Q(\stackrel{鈫�}{E}+\stackrel{鈫�}{v}脳\stackrel{鈫�}{B})$...... (1)

System A;

Substitute$+q$for$Q$, $\frac{1}{4蟺{蔚}_o}\frac{鈭�q}{d^2}\stackrel{鈭�}{y}$for$\stackrel{鈫�}{E}$,$鈭�v\left(\stackrel{鈭�}{x}\right)$for$\stackrel{鈫�}{v}$ and $\frac{{渭}_o}{4蟺}\left(\frac{qv}{d^2}\right)\stackrel{鈭�}{z}$for $\stackrel{鈫�}{B}$in the equation (1).

$\begin{array}{c}\stackrel{鈫�}{F}=q\left(\frac{1}{4蟺{蔚}_o}\frac{鈭�q}{d^2}\stackrel{鈭�}{y}\right)+鈭�v\left(\stackrel{鈭�}{x}\right)脳\frac{{渭}_o}{4蟺}\left(\frac{qv}{d^2}\right)\stackrel{鈭�}{z}\\ =\left(\frac{1}{4蟺{蔚}_o}\frac{鈭�q^2}{d^2}\stackrel{鈭�}{y}\right)+\left(\frac{{渭}_o}{4蟺}\left(\frac{q^2{v}^2}{d^2}\stackrel{鈭�}{y}\right)\right)\end{array}$

System B;

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

Substitute $+q$for $Q$,$\frac{1}{4蟺{蔚}_o}\frac{鈭�q}{d^2}\stackrel{鈭�}{y}$for $\stackrel{鈫�}{E}$,$0$for $\stackrel{鈫�}{v}$and $\frac{{渭}_o}{4蟺}\left(\frac{qv}{d^2}\right)\stackrel{鈭�}{z}$for$\stackrel{鈫�}{B}$in the equation (1).

role="math" localid="1658133092641" $\begin{array}{c}\stackrel{鈫�}{F}=q\left(\frac{1}{4蟺{蔚}_o}\frac{鈭�q}{d^2}\stackrel{鈭�}{y}\right)+鈭�0脳\frac{{渭}_o}{4蟺}\left(\frac{qv}{d^2}\right)\stackrel{鈭�}{z}\\ =\left(\frac{1}{4蟺{蔚}_o}\frac{鈭�q^2}{d^2}\stackrel{鈭�}{y}\right)\end{array}$

System C;

The velocity of the charge$鈭�q$is zero.

Substitute $+q$for $Q$, $\frac{1}{4蟺{蔚}_o}\frac{鈭�q}{d^2}\stackrel{鈭�}{y}$for $\stackrel{鈫�}{E}$, $v(鈭�\stackrel{鈭�}{x})$for $\stackrel{鈫�}{v}$ and for $\stackrel{鈫�}{B}$in the equation (1).

.$\begin{array}{c}\stackrel{鈫�}{F}=q\left(\frac{1}{4蟺{蔚}_o}\frac{鈭�q}{d^2}\stackrel{鈭�}{y}\right)+v(鈭�\stackrel{鈭�}{x})脳0\\ =\left(\frac{1}{4蟺{蔚}_o}\frac{鈭�q^2}{d^2}\stackrel{鈭�}{y}\right)\end{array}$

Therefore,