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Physics Lesson 16.3.3 - Magnetic Force between two Parallel Current Carrying Conducting Wires
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Welcome to our Physics lesson on Magnetic Force between two Parallel Current Carrying Conducting Wires, this is the third lesson of our suite of physics lessons covering the topic of Magnetic Force on a Current Carrying Wire. Ampere's Force, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.
Magnetic Force between two Parallel Current Carrying Conducting Wires
Since the interaction between magnetic field and electric current brings a magnetic force on the wire, we will have a mutual interaction if two current carrying wires are placed inside a magnetic field. The simplest case in this regard involves two parallel current carrying conductors inside a magnetic field where currents lie in the same direction as shown in the figure.
The two parallel wires having lengths L1 and L2 respectively are separated by a distance d from each other. In addition, the magnetic fields they produce are B1 and B2 respectively. As a result, a magnetic force F12 produced by the first wire will act on the second wire while the magnetic force F21 produced by the second wire acts on the first one. These forces are equal and opposite based on the action-reaction principle (Newton's Third Law of Motion). You can also apply the Fleming's Left Hand Rule to convince yourself that these forces have opposite directions.
As for the magnetic fields, we can use the following reasoning:
The first wire (which carries a current of I1) produces a magnetic field B12 at the position of the second wire (at distance d from the first wire). The magnitude of B12 produced by the first wire at the position of the second wire therefore is
and the corresponding magnetic force F12 of the first wire on the second, is
= μ0 ∙ I1 ∙ I2/(2π ∙ d) ∙ L2
Likewise, the magnetic force F21 by which the second wire acts on the first, is
= μ0 ∙ I1 ∙ I2/(2π ∙ d) ∙ L1
If the two wires have the same length and current, the magnitudes of the two above forces are equal. Since forces have opposite directions, the wires repel each other when parallel currents flow in them. On the other hand, when currents are antiparallel, i.e. when they have opposite directions, the two wires attract each other based on the direction of magnetic forces produced. Look at the figure:
Example 3
Two identical parallel bars are placed horizontally one above the other as shown in the figure.
The bars are 50 cm long each. The lower bar carries a current of 800A and it is unmoveable, while the upper bar (the 500A bar) can move freely when released and it is 20 cm above the lower bar. What is the mass of the upper bar so that it will stay in the actual position after being released? Take g = 9.81 m/s2.
Solution 3
Clues:
I1 = 800A = 8 × 102 A
I2 = 500A = 5 × 102 A
L1 = L2 = 50 cm = 0.50 m
d = 20 cm = 0.20 m
(g = 9.81 m/s2)
(μ0 = 4π × 10-7 N/A2)
m2 = ?
The condition for the upper bar to stay unmoveable in the actual position after release is that magnetic force by which the lower bar acts on it, be equal to the weight of the upper bar, i.e.
μ0 ∙ I1 ∙ I2/2π ∙ d ∙ L2 = m2 ∙ g
m2 = μ0 ∙ I1 ∙ I2 ∙ L2/2π ∙ d ∙ g
= (4 ∙ 3.14 × 10-7 N/A2 ) ∙ (8 × 102 A) ∙ (5 × 102 A) ∙ (0.50m)/2 ∙ 3.14 ∙ (0.20m) ∙ (9.81 m/s2)
= 20.4 × 10-3 kg
= 20.4 g
You have reached the end of Physics lesson 16.3.3 Magnetic Force between two Parallel Current Carrying Conducting Wires. There are 5 lessons in this physics tutorial covering Magnetic Force on a Current Carrying Wire. Ampere's Force, you can access all the lessons from this tutorial below.
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