Question

Two infinitely long parallel wires having linear charge densities ${\lambda _1}$ and ${\lambda _2}$ respectively are placed at a distance of $R$ meters. The force per unit length on either wire will be $\left( {K = \dfrac{1}{{4\pi {\varepsilon _0}}}} \right)$
A. $K\dfrac{{2{\lambda _1}{\lambda _2}}}{{{R^2}}}$
B. $K\dfrac{{2{\lambda _1}{\lambda _2}}}{R}$
C. $K\dfrac{{{\lambda _1}{\lambda _2}}}{{{R^2}}}$
D. $K\dfrac{{{\lambda _1}{\lambda _2}}}{R}$

Answer 1

Here, we will use the formula of the electric field to calculate the force per unit length on either wire. For calculating the charge on the wire, we will first consider a small charge in the wire, then we will calculate the charge on the whole wire by integrating the equation. Also, the wires are separated by a distance of $R$ .

Complete step by step answer:
Consider two infinitely long parallel wires that are separated by a distance $R$ . Therefore, the electric field on the wire having linear charge density ${\lambda _1}$ is given by
$E = \dfrac{{{\lambda _1}}}{{2\pi {\varepsilon _0}R}}$
Now, let the wires carry current $I$ such that there will be the movement of charges in the wire. Therefore, the charge density will be ${\lambda _1}$. Hence, the charge on the wire will be given by
$dQ = {\lambda _1}dl$
Therefore, the total charge in the wire can be calculated by integrating the above equation between the limits $0$ to $l$ as shown below
$\int {dQ} = \int\limits_0^l {{\lambda _1}dl}$
$\Rightarrow \,{Q_1} = {\lambda _1}\int\limits_0^l {dl}$
$\Rightarrow \,{Q_1} = {\lambda _1}l$
Similarly for second wire, the charge density will be ${\lambda _2}$, hence, the charge on the second wire will be
${Q_2} = {\lambda _2}l$
Now, the force can be calculated as given by
$F = QE$
Therefore, the force on the second wire can be calculated as
$F = {Q_2}E$
Now, putting the values of $Q$ and $E$ in the above equation, we get
$\dfrac{F}{l} = {\lambda _2}l \times \dfrac{{{\lambda _1}}}{{2\pi {\varepsilon _0}R}}$
$\therefore\dfrac{F}{l} = K\dfrac{{2{\lambda _1}{\lambda _2}}}{R}$
Hence, the force per unit length on either wire will be $K\dfrac{{2{\lambda _1}{\lambda _2}}}{R}$ .

Hence, the option B is the correct option.

Note: Here, the force per unit length on both the wires will be the same. For this you can take the electric field having linear charge density ${\lambda _2}$ . Putting this value in the formula of force on the wire carrying charge ${Q_1}$ , we will get the same value of force.