Question

A2.50-msegment of wire supplying current to the motor of a submerged submarine carries 1000 A and feels a 4.00-Nrepulsive force from a parallel wire 5.00cmaway. What is the direction and magnitude of the current in the other wire?

Short answer

The current magnitude in the wire is 400 A and direction will be opposite to the other wire.

Step-by-step solution

Given data

The force between two wires is $\mathrm{F}=4.00\mathrm{N}$

The separation between the wires is $\mathrm{r}=5.00\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=5.00\times {10}^{-2}\mathrm{m}$

Current in the first wire is ${\mathrm{I}}_1=1000\mathrm{A}$

Length of the wire segment$\mathrm{L}=2.50\mathrm{cm}$

Define current

A flow of electrical charge carriers, generally electrons or electron-deficient atoms, is referred to as current.

Evaluating the magnitude and direction of the current

To determine the current in another wire, we will use the formula for force per unit length that can be expressed as,

$\frac{\mathrm{F}}{\mathrm{L}}=\frac{{\mathrm{\mu }}_0{\mathrm{I}}_1{\mathrm{I}}_2}{2\mathrm{\tau \tau r}}$……………(1)

Where ${\mathrm{I}}_1$and ${\mathrm{I}}_2$is the current in the wires, r is the separation between the wires, and ${\mathrm{\mu }}_0$is the permeability of free space.

Substitute the given data in equation (1) and solve for ${\mathrm{I}}_2$to get the current in another wire, such that

$$\begin{gathered} I_2=\frac{2\mathrm{\pi rF}}{{\mu }_0{I}_{1}L} \\ =\frac{2\mathrm{\pi }\times 4.00\mathrm{N}\times 5.00\times {10}^{-2}\mathrm{m}}{4\mathrm{\pi }\times {10}^{-7}\mathrm{N}/{\mathrm{A}}^2\times 1000\mathrm{A}\times 2.50\mathrm{m}} \\ =400\mathrm{A} \end{gathered}$$

Since the force is repulsive, the current in the other wire will flow in the opposite direction as the current in the first.

Therefore, the current magnitude is 400 A and direction is opposite of the other wire.