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Q52PE

Expert-verifiedFound in: Page 813

Book edition
1st Edition

Author(s)
Paul Peter Urone

Pages
1272 pages

ISBN
9781938168000

**A **** 2.50-m ****segment of wire supplying current to the motor of a submerged submarine carries 1000 A and feels a 4.00-N**** repulsive force from a parallel wire 5.00cm**** away. What is the direction and magnitude of the current in the other wire?**

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

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 $\mathrm{L}=2.50\mathrm{cm}$

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

${I}_{2}=\frac{2\mathrm{\pi rF}}{{\mu}_{0}{I}_{1}L}\phantom{\rule{0ex}{0ex}}=\frac{2\mathrm{\pi}\times 4.00\mathrm{N}\times 5.00\times {10}^{-2}\mathrm{m}}{4\mathrm{\pi}\times {10}^{-7}\mathrm{N}/{\mathrm{A}}^{2}\times 1000\mathrm{A}\times 2.50\mathrm{m}}\phantom{\rule{0ex}{0ex}}=400\mathrm{A}$

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.

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