Question

Suppose you have two infinite straight line charges$\lambda$, a distance d apart, moving along at a constant speed $\upsilon$(Fig. 5.26). How great would have tobe in order for the magnetic attraction to balance the electrical repulsion? Work out the actual number. Is this a reasonable sort of speed?

Short answer

The required speed is$\upsilon =3\times {10}^{8}m/s$ and it is not reasonable two wires cannot attain that speed.

Step-by-step solution

Define function

Write the expression for the magnitude of force between two parallel straight current carrying conductors.

$F=\frac{{\mu }_0{i}_1{i}_2}{2\pi d}$ …… (1)

Here,$i_1$ and $i_2$are currents through the parallel conductors,d is the distance between the two parallel conductors and ${\mu }_0$is the permeability.

Write the expression for the electric filed at distance of due to infinite straight conducting wire.

$E=\frac{\lambda }{2\pi {\epsilon }_{0}d}$ ……. (2)

Here,$\lambda$ is the linear charge density and ${\epsilon }_0$is the permittivity.

Determine speed

A line charge moving with speed constitute current that is,

$i_1=i_2=\lambda \upsilon$

Now, let’s consider that, one of the wires, then write the expression for the attractive magnetic force on his wire per unit length.

$F=\frac{{\mu }_0{i}_1{i}_2}{2\pi d}$

Substitute $\lambda \upsilon$for $i_1$also $\lambda \upsilon$for $i_2$.

$\begin{array}{rcl}F& =& \frac{{\mu }_0\left(\lambda \upsilon \right)\left(\lambda \upsilon \right)}{2\pi d}\\ & =& \frac{{\mu }_0{\lambda }^2{\upsilon }^2}{2\pi d}\\ & & \end{array}$

Write the expression for electrostatic repulsion per unit length.

$F_C=\frac{{\lambda }^2}{2\pi {\epsilon }_{0}d}$

The line charges have to balance the magnetic attraction and electrical repulsion.

$F_C=F$ …… (3)

Substitute $\frac{{\lambda }^2}{2\pi {\epsilon }_{0}d}$for $F_C$and$\frac{{\mu }_0{\lambda }^2{\upsilon }^2}{2\pi d}$for F in equation (3),

$$\begin{gathered} \frac{{\lambda }^2}{2\pi {\epsilon }_{0}d}=\frac{{\mu }_0{\lambda }^2{\upsilon }^2}{2\pi d} \\ \upsilon =\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}} \end{gathered}$$

Substitute $4\pi \times {10}^{-7}H/M$for${\mu }_0$and $8.85\times {10}^{-12}{C}^2/N·m^2$for ${\epsilon }_0$in above equation.

$$\begin{gathered} \upsilon =\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}} \\ \upsilon =\frac{1}{\sqrt{\left(4\pi \times {10}^{-7}H/M\right)}\left(8.85\times {10}^{-12}{C}^2/N.m^2\right)} \\ \upsilon =3\times {10}^{8}m/s \end{gathered}$$

Thus, the required speed is $\upsilon =3\times {10}^{8}m/s$and it is not reasonable two wires cannot attain that speed.