Two parallel wires, separated by a perpendicular distance of 5.0 cm, carry currents of 5.0 A and 2.0 A in the same direction. What force does the 5.0-A wire exert on a 50 cm-length of the 2.0-A wire? a. 60 μN, repulsive b. 20 μN, attractive c. 100 μN, attractive d. 80 μN, attractive e. 40 μN, attractive
Two parallel wires, separated by a perpendicular distance of 5.0 cm, carry currents of 5.0 A and 2.0 A in the same direction. What force does the 5.0-A wire exert on a 50 cm-length of the 2.0-A wire? a. 60 μN, repulsive b. 20 μN, attractive c. 100 μN, attractive d. 80 μN, attractive e. 40 μN, attractive
College Physics
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Chapter1: Units, Trigonometry. And Vectors
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Two parallel wires, separated by a perpendicular distance of 5.0 cm, carry currents of 5.0 A and 2.0 A in the same direction. What force does the 5.0-A wire exert on a 50 cm-length of the 2.0-A wire?
a. 60 μN, repulsive
b. 20 μN, attractive
c. 100 μN, attractive
d. 80 μN, attractive
e. 40 μN, attractive
![**Question:**
Two parallel wires, separated by a perpendicular distance of 5.0 cm, carry currents of 5.0 A and 2.0 A in the same direction. What force does the 5.0-A wire exert on a 50 cm-length of the 2.0-A wire?
**Options:**
a. 60 μN, repulsive
b. 20 μN, attractive
c. 100 μN, attractive
d. 80 μN, attractive
e. 40 μN, attractive
**Explanation:**
This problem involves calculating the magnetic force between two parallel current-carrying wires. The magnetic force per unit length between two parallel wires separated by a distance \(d\) and carrying currents \(I_1\) and \(I_2\) is given by:
\[
F/L = \frac{{\mu_0 \cdot I_1 \cdot I_2}}{{2\pi d}}
\]
where \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{N/A}^2\)).
To find the total force on the 50 cm (0.5 m) length of the wire:
\[
F = (F/L) \cdot \text{length}
\]
When the currents are in the same direction, the force is attractive.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ede0637-a731-4aa5-a543-54dacf89e51e%2F44d4bcef-f25e-4609-847c-7c81b4fd981f%2F079cbsj_processed.png&w=3840&q=75)
Transcribed Image Text:**Question:**
Two parallel wires, separated by a perpendicular distance of 5.0 cm, carry currents of 5.0 A and 2.0 A in the same direction. What force does the 5.0-A wire exert on a 50 cm-length of the 2.0-A wire?
**Options:**
a. 60 μN, repulsive
b. 20 μN, attractive
c. 100 μN, attractive
d. 80 μN, attractive
e. 40 μN, attractive
**Explanation:**
This problem involves calculating the magnetic force between two parallel current-carrying wires. The magnetic force per unit length between two parallel wires separated by a distance \(d\) and carrying currents \(I_1\) and \(I_2\) is given by:
\[
F/L = \frac{{\mu_0 \cdot I_1 \cdot I_2}}{{2\pi d}}
\]
where \(\mu_0\) is the permeability of free space (\(4\pi \times 10^{-7} \, \text{N/A}^2\)).
To find the total force on the 50 cm (0.5 m) length of the wire:
\[
F = (F/L) \cdot \text{length}
\]
When the currents are in the same direction, the force is attractive.
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