Question

Two long parallel wires are separated by a distance of $2.50\mathrm{cm}$. The force per unit length that each wire exerts on the other is $4.00\times {10}^{-5}\mathrm{N}/\mathrm{m}$, and the wires repel each other. The current in one wire is $0.600\mathrm{A}$.

(A). What is the current in the second wire?

(B). Are the two currents in the same direction or in opposite directions?

Answer 1

Step 1: Given parameters

Two long parallel wires separated by a distance, $r=2.50\mathrm{cm}$,

Force per unit length exerted by each wire, $\frac{F}{l}=4.0\times {10}^{-5}\mathrm{N}/\mathrm{m}$.

Current in one wire, $i_1=0.600\mathrm{A}$.

Step 2: Calculate the current in the second wire

(A)

Calculate the magnetic force per unit length between two parallel conductors carrying current by using the expression given as:

$\frac{F}{l}=\mu \frac{i_1\times i_2}{2\pi r}$

Where, $\mu$ is the permeability of free space and is equal to $4\pi \times {10}^{-7}\mathrm{T}{\mathrm{mA}}^{-1}$, $\frac{F}{l}$ is the force per unit length between two parallel currents $i_1$ and $i_2$ separated by a distance $r$.

The current in the secondary wire($i_2$) is given by,

$\frac{F}{l}=\mu \frac{i_1\times i_2}{2\pi r}$

Substitute the values in the above equation,

$\begin{array}{rcl}4.00\times {10}^{-5}& =& 4\pi \times {10}^{-7}\frac{0.600\times i_2}{2\pi \times 2.5\times {10}^{-2}}\\ & =& 0.48\times {10}^{-5}\times i_2\end{array}$

So,

$\begin{array}{rcl}{i}_2& =& \frac{4\times {10}^{-5}}{0.48\times {10}^{-5}}\\ & =& 8.33\mathrm{A}\end{array}$

(B)

1. When the currents in the wire are in the same direction, they experience an attractive force and when they carry current in opposite directions, they experience a repulsive force.
2. As two parallel current-carrying wires repel each other which implies that the current ($i_1$ and $i_2$) are in the opposite direction.
3. Hence, the currents in the wire are in the opposite direction.