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College Physics (Urone)
Found in: Page 813

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Short Answer

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.

See the step by step solution

Step by Step Solution

Step 1: Given data

The force between two wires is F=4.00 N

The separation between the wires is r=5.00 cm1m100 cm=5.00×10-2 m

Current in the first wire is I1=1000 A

Length of the wire segment L=2.50 cm

Step 2: Define current

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

Step 3: 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,

FL=μ0I1I22ττr ……………(1)

Where I1 and I2 is the current in the wires, r is the separation between the wires, and μ0 is the permeability of free space.

Substitute the given data in equation (1) and solve for I2 to get the current in another wire, such that

I2=2πrFμ0I1L =2π×4.00 N×5.00×10-2 m4π×10-7 N/A2×1000 A×2.50 m =400 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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