Mastering Physics Solutions: Exercise 19.55

Mastering Physics Solutions: Exercise 19.55

On May 14, 2012, in Chapter 19: Magnetism, by Mastering Physics Solutions

Exercise 19.55

Part A = repulsive
Part B = Since the currents flow in opposite directions, then due to the right hand rule, the magnetic field induced by one wire will be in an opposite direction to that of the other wire.
Part C = 10 μT Click to use the calculator/solver for this part of the problem
Part D = 50 μT/m Click to use the calculator/solver for this part of the problem

Solutions Below:

Two long, straight, parallel wires 10 cm apart carry currents in opposite directions.

Part A

Use the right-hand source and force rules to determine whether the forces on the wires are attractive or repulsive.

Currents in opposite directions creates a repulsive field, currents in the same direction create an attractive field.

repulsive

Part B

Show your reasoning.

Since the currents flow in opposite directions, then due to the right hand rule, the magnetic field induced by one wire will be in an opposite direction to that of the other wire.

Part C

If the wires carry equal currents of 5.0 A, what is the magnetic field magnitude that each produces at the other’s location?
Express your answer using two significant figures.

The magnetic field induced by current flowing through a straight wire is:

B = ((μ0 * I) / (4 * π * R)) * (sin(φ1) + sin(φ2))

Your class may or may not use this general formula, but φ1 and φ2 are just the angles above and below a line perpendicular to the straight wire. Basically, the straight wire gets divided into an infinite number of segments and the magnetic field from each segment is added up (through calculus) to reach a total. The φ’s are just the angles between each segment and a line that runs perpendicular between the wire and the point where the magnetic field is being measured. Since the wires in this problem can be assumed to be infinitely long (the problem doesn’t say otherwise), the φ’s both become π / 2, and the formula becomes:

B = ((μ0 * I) / (4 * π * R)) * (sin(π / 2) + sin(π / 2))

So:

B = ((μ0 * I) / (2 * π * R))

Therefore:

B = (1.2566*10^-6 * 5) / (2 * 3.14159 * 0.1)
B = 10 μT

10 μT

Part D

Use the result of part C to determine the magnitude of the force per unit length they exert on each other.
Express your answer using two significant figures.

The force from one wire on the other is equal to:

F = ILB

And since B was found to be 10 μT in Part C:

F = ILB
F = 5 * L * 0.000010

For length, use 1 m since this is a unit length problem:

F = 5 * 1 * 0.000010
F = 50 μT/m

This is sort of misleading because the 50 μT/m is just from one wire – the actual repulsive force is double this since there are two wires, but mastering physics just wants the force from 1 wire (50 μT/m)

50 μT/m

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