Lesson Explainer: Electromagnetic Interactions between Straight Conductors Physics

In this explainer, we will learn how to determine the magnetic forces between parallel lines of current and analyze the net magnetic fields of the currents.

When charge flows along a wire, it creates a magnetic field around the wire, as shown below.

The field points along concentric circles centered on the wire, as follows, with the strength of the field being inversely proportional to the distance from the wire.

The direction of the magnetic field is determined by the direction of the current. Pointing the thumb of our right hand in the current direction, our fingers curl closed in the direction of the magnetic field.

If two current-carrying wires are parallel, they interact through their magnetic fields. Though limited in the figure below, these fields extend infinitely far from each wire.

This interaction has two effects. One is to create a net magnetic field due to the two wires. The second is to generate forces experienced by each wire.

Consider first the net magnetic field. Due to each wire, there is a magnetic field vector at each point in space.

For illustration purposes, we can choose a point equidistant from the wires, as shown below.

At point P, there is a magnetic field vector due to current 𝐼 and a magnetic field vector due to current 𝐼, as follows.

The net magnetic field at P, labeled 𝐡net below, is the vector sum of the two magnetic fields at this point.

Example 1: Ranking Magnetic Field Strengths from Current-Carrying Wires

The diagram shows concentric field lines of the magnetic fields of two parallel current-carrying conductors. The right-hand current goes into the plane of the diagram and the left-hand current goes out of the plane of the diagram. The currents have the same magnitude as each other. The increase in the radius of the concentric field lines is constant, and the strength of a magnetic field at a point around a current is inversely proportional to the perpendicular distance of the point from the current.

Which of the following lists of the points shown in the diagram correctly orders the points from the greatest to the smallest magnitude of net magnetic field?

  1. 𝐢, 𝐷, 𝐸, 𝐡, 𝐴
  2. 𝐴, 𝐸, 𝐢, 𝐡, 𝐷
  3. 𝐸, 𝐡, 𝐴, 𝐢, 𝐷
  4. 𝐴, 𝐡, 𝐸, 𝐢, 𝐷
  5. 𝐷, 𝐸, 𝐢, 𝐡, 𝐴

Answer

We begin by indicating the relative strengths of the magnetic fields due to each current-carrying wire at each of the five points of interest.

Consider first the current on the left, which points out of the screen toward us. Pointing the thumb of our right hand out of the screen in the direction of this current, our fingers curl counterclockwise.

Therefore, at each point along a ring centered on this wire, the magnetic field due to this current will point tangent to the ring in the counterclockwise direction.

The strength of the field at any point is determined by the distance from the wire axis. Field strength is inversely proportional to distance, so the field at a point twice as far from the wire will be half as strong. We can indicate these directions and relative strengths for the current we will call 𝐼 on the diagram below.

Next, we consider the magnetic field created by the second current, which we will call 𝐼.

Pointing our right thumb into the screen in the direction of 𝐼, our fingers curl clockwise. Therefore, the magnetic field due to this current will point tangent to any circle centered on the wire in the clockwise direction.

At points 𝐴, 𝐡, 𝐢, 𝐷, and 𝐸, those relative field strengths and directions appear as follows.

Now, we solve for the net magnetic field strength at each of these points by adding the fields as vectors. We do this using the tip-to-tail method of vector addition, as shown below.

Adding these vectors together using this method, we find the following resultant vectors at each point of interest, shown in the following diagram.

From greatest to smallest net magnetic field strength, we see these points are ordered as 𝐢, 𝐷, 𝐸, 𝐡, 𝐴. This corresponds to option 𝐴.

If the two magnetic fields are added together at every point in space, the net magnetic field due to 𝐼 and 𝐼 emerges, as shown below.

We may superimpose the results from the previous example onto this diagram. The results appear as follows.

Note that this field is created by currents moving antiparallel to one another and that the density of field lines indicates the strength of the magnetic field.

Along with the net field due to antiparallel currents, we consider the net magnetic field generated by two currents pointing in the same direction, as shown below.

In this instance, the magnetic field directly between the two wires is zero, and the field in that general area is weak.

Example 2: Determining Current Directions from Net Magnetic Field

Several horizontal pairs of parallel conducting wires are stacked vertically. The magnitude of the current in each wire is the same. A cross section of the resultant magnetic field due to the currents is shown in the diagram. Which of the configurations of current directions shown would produce the resultant magnetic field?

  1. III
  2. I
  3. I and III
  4. I and IV
  5. III and IV

Answer

Studying the magnetic field diagram, we recall that magnetic field strength is indicated by the density of field lines; the greater the density, the stronger the field.

If we consider a vertical line midway between each horizontal pair of current-carrying wires, as shown below, we see that field lines do exist along this line, but they do not act in a consistent direction.

If there were no magnetic field lines along this vertical orange line, then the total field along it would be zero.

A magnetic field of zero between a pair of current-carrying wires occurs when the wires carry current in the same direction. Since we are not seeing that in this example, we can conclude that configuration II will not be part of our final answer. For such a configuration, the magnetic field along a line down the middle would be zero, but here we see it is not.

Looking next at configuration III, we see all the wires in the left column carry current pointing into the screen, while those on the right carry current pointing out of it.

We have seen before that a single pair of wires carrying currents in opposite directions creates a magnetic field with densely concentrated field lines pointing vertically between the pair of wires.

Stacking many such pairs as in configuration III will amplify this effect, essentially creating the magnetic field that appears inside a solenoid.

Because the actual magnetic field in this example does not show a strong, vertically oriented field between the pairs of wires, we can say this field cannot be due to configuration III.

Configurations I and IV remain. Each of these involves alternating pairs of oppositely directed currents.

Regarding these pairs, when the current in the left wire points out of the screen and that in the right wire points into it, the magnetic field between that pair of wires points vertically upward.

When the current directions are reversed in the two wires, the field between them remains oriented vertically but points downward.

Therefore, the fields between alternating pairs of wires in configurations I and IV will oppose one another. This gives rise to the pattern of closed loops around each wire that we see in the magnetic field diagram.

Note that the diagram does not indicate the direction of the field in any of the loops; it could be clockwise or counterclockwise.

For these reasons, both configurations I and IV could generate the overall magnetic field seen in this diagram. For our answer, we choose option D: configurations I and IV.

We now consider the forces acting on a pair of parallel current-carrying wires.

Recall that a current 𝐼 will create a magnetic field around itself. The magnitude of this field is given by the equation 𝐡=πœ‡πΌ2πœ‹π‘Ÿ, where πœ‡οŠ¦ is a constant called the permeability of free space, 𝐼 is the magnitude of the current generating the field, and π‘Ÿ is the distance from the axis of the current at which the field strength is being calculated.

Consider two parallel wires of length 𝐿 carrying currents 𝐼 and 𝐼 and separated by a distance 𝑑, as shown below.

We can recall the general relationship for the force between these wires: 𝐹=𝐡𝐼𝐿.

If we consider the force on the wire carrying current 𝐼, our force equation becomes 𝐹=𝐡𝐼𝐿.

Here, 𝐹 is the force on the wire carrying current 𝐼, 𝐡 is the magnetic field at that wire due to current 𝐼, and 𝐿 is the wire length.

Since 𝐡=πœ‡πΌ2πœ‹π‘‘, we can substitute the right side of this equation into our equation for 𝐹: 𝐹=ο€½πœ‡πΌ2πœ‹π‘‘ο‰Γ—πΌΓ—πΏ.

Dividing both sides by 𝐿 and grouping terms differently, 𝐹𝐿=πœ‡πΌπΌ2πœ‹π‘‘.

This is the equation for the force per unit length on the wire carrying current 𝐼. Note that this force depends on the magnitudes of both currents.

If we considered instead the force on the wire carrying current 𝐼, the result for 𝐹𝐿 would be as follows: 𝐹𝐿=πœ‡πΌπΌ2πœ‹π‘‘.

𝐹 and 𝐹 are equal in magnitude and opposite in direction.

Example 3: Solving for the Distance between Two Current-Carrying Wires

Two long, straight, parallel conducting wires are separated by distance 𝑑, as shown in the diagram. The wires both carry currents of 1.6 A in the same direction. A length 𝐿=0.75m of each wire exerts a force of 3.5 ΞΌN on the other. Find the distance 𝑑. Use a value of 4πœ‹Γ—10 H/m for the magnetic permeability of the region between the wires. Give your answer to two decimal places.

Answer

We can write an equation for the force on either current-carrying wire as follows: 𝐹=πœ‡πΌπΌπΏ2πœ‹π‘‘.

Here. it is the distance 𝑑 between the wires, rather than the force, that we want to solve for.

We can rearrange to make 𝑑 the subject of this equation by multiplying both sides by 𝑑𝐹: 𝑑=πœ‡πΌπΌπΏ2πœ‹πΉ.

We are told that the permeability of free space, πœ‡οŠ¦, is 4πœ‹Γ—10 H/m. H stands for henries, the SI unit of inductance.

Currents 𝐼 and 𝐼 are both 1.6 A, while 𝐿 is given as 0.75 m, and the force, 𝐹, is 3.5 ΞΌN, or 3.5Γ—10 N.

Substituting in these values, 𝑑=ο€Ή4πœ‹Γ—10/×(1.6)Γ—(1.6)Γ—(0.75)2πœ‹Γ—(3.5Γ—10)=0.11.HmAAmNm

Rounded to two significant figures, the distance between the wires is 0.11 m.

Given two parallel current-carrying wires, it is possible to calculate not only the magnitude of the force acting on each wire, but also its direction.

To do this for a certain wire, we consider only the magnetic field the wire experiences (rather than the field it creates) and the current in the wire.

The force direction on a current-carrying wire in a magnetic field is determined using the right-hand rule, as shown below.

First, we point the first finger of our right hand in the direction of the current in the wire of interest.

Next, we point our second finger in the direction of the magnetic field experienced by the wire.

When our right thumb points perpendicularly to both our first and second fingers, it shows the direction of the force acting on the wire.

Let’s now apply this method to find the directions of the forces acting on two current-carrying wires.

Suppose two straight wires, π‘ŠοŠ§ and π‘ŠοŠ¨, carrying currents 𝐼 and 𝐼 are parallel to one another, as shown below.

Based on the right-hand rule for determining the direction of the magnetic field created by each current, we know that the field at π‘ŠοŠ§ due to 𝐼 will point out of the screen, as shown.

To determine the direction of the magnetic force on π‘ŠοŠ§, first, we point the first finger of our right hand in the direction of the magnetic field 𝐡 at π‘ŠοŠ§ (out of the screen). Then, we point our second finger in the direction of the current in π‘ŠοŠ§ (to the right, as per 𝐼).

The direction our right thumb faces shows the direction of the magnetic force on π‘ŠοŠ§, depicted below.

We can determine the direction of the force on π‘ŠοŠ¨ using the same method.

The magnetic field direction at π‘ŠοŠ¨ due to 𝐼 is into the screen, as indicated below.

With our right forefinger pointing in the direction of 𝐼 and our second finger pointing into the screen with 𝐡, our right thumb points vertically upward, the direction of the force on π‘ŠοŠ¨.

It is generally true that when parallel wires carry currents in the same direction, the forces on the wires are attractive, as shown below.

Note that if the magnitudes of the currents 𝐼 and 𝐼 are the same, the net magnetic field midway between the two wires is zero.

Consider an alternate scenario, shown below, where 𝐼 and 𝐼 point in opposite directions. The forces now act to push the wires apart.

In this instance, the net magnetic field between the two wires is not zero; instead of canceling one another, the fields both point in the same direction.

Example 4: Determining the Net Force on a Current-Carrying Conductor

Three long, straight, parallel, conducting wires π‘ŠοŠ§, π‘ŠοŠ¨, and π‘ŠοŠ© carry currents of 1.6 A, 1.1 A, and βˆ’1.9 A respectively. π‘ŠοŠ§ is 2.5 cm from π‘ŠοŠ¨ and 5.1 cm from π‘ŠοŠ©. π‘ŠοŠ¨ is located between the other two wires. Find the magnitude of the force per meter of length on π‘ŠοŠ§ perpendicular to π‘ŠοŠ¨. Use a value of 4πœ‹Γ—10 H/m for the magnetic permeability of the region between the wires. Give your answer in scientific notation to one decimal place.

Answer

We can sketch the relative positions of the three wires as follows.

The force per meter of length on π‘ŠοŠ§ will be due to forces from π‘ŠοŠ¨ and π‘ŠοŠ©. We will first determine the directions of those forces before combining them as vectors.

Considering the force on π‘ŠοŠ§ due to 𝐼, we can use the right-hand rule to discover that the magnetic field direction at π‘ŠοŠ§ due to 𝐼 is out of the screen, as shown.

We then apply the specific right-hand rule we learned above for determining the direction of the force on π‘ŠοŠ§ due to 𝐼.

Pointing our right forefinger in the direction of 𝐼 and our second finger out of the screen in the direction of 𝐡, our thumb points downward, from π‘ŠοŠ§ toward π‘ŠοŠ¨. The following figure shows this force on π‘ŠοŠ§ due to the current in π‘ŠοŠ¨.

Following a similar process to determine the direction of the force on π‘ŠοŠ§ due to 𝐼, we find that the magnetic field at π‘ŠοŠ§ due to 𝐼 points into the screen.

This means the force on π‘ŠοŠ§ from 𝐼 points upward on our diagram, away from π‘ŠοŠ¨, as follows.

We see that the two forces on π‘ŠοŠ§ act in opposite directions. In the diagram above, the arrows representing forces are to be compared in direction but not in length; we do not yet know which force is stronger.

We do know, however, that when we sum these forces, one will be positive and the other negative.

To compute the net force on π‘ŠοŠ§, we recall that for two parallel wires carrying currents 𝐼 and 𝐼, the force per unit length on the wire carrying 𝐼 is given by 𝐹𝐿=πœ‡πΌπΌ2πœ‹π‘‘.

In this example, we want to calculate what we will represent as 𝐹𝐿, the net force on π‘ŠοŠ§ per metre of length. Since there are two forces combining to make the total force, 𝐹𝐿=πœ‡πΌπΌ2πœ‹π‘‘+πœ‡πΌπΌ2πœ‹π‘‘, where π‘‘οŠ¨ (2.5 cm) and π‘‘οŠ© (5.1 cm) are the distances between π‘ŠοŠ§ and π‘ŠοŠ¨ and between π‘ŠοŠ§ and π‘ŠοŠ© respectively.

We can factor several values out of the terms on the right so that 𝐹𝐿=πœ‡πΌ2πœ‹Γ—ο€½πΌπ‘‘+𝐼𝑑.

The permeability in the space around the wires is given as 4πœ‹Γ—10 H/m, which equals πœ‡οŠ¦.

The three currents are also given. And we note that 𝐼 is negative, while 𝐼 and 𝐼 are positive. The difference in sign between 𝐼 and 𝐼 accounts for the difference in direction between forces 𝐹 and 𝐹.

The distances π‘‘οŠ¨ and π‘‘οŠ© are known, but note that they are given in units of centimetres. For the units across all the values we are using to agree, we need to convert these distances into units of metres, the SI base unit of distance.

Since there are 100 centimetres in 1 metre, we can write that 𝑁=𝑁×10,cmm where 𝑁 is a number of centimetres.

To convert π‘‘οŠ¨ and π‘‘οŠ© to distances in metres, we multiply each value by 10: 𝑑=2.5=2.5Γ—10=0.025,𝑑=5.1=5.1Γ—10=0.051.cmmmcmmm

We can now substitute these in and solve for the force on π‘ŠοŠ§ per metre: 𝐹1=ο€Ή4πœ‹Γ—10/×(1.6)2πœ‹Γ—ο€Ό1.10.025+βˆ’1.90.051=2.158431…×10/.mHmAAcmAmNm

Giving our answer in scientific notation to one decimal place, the magnitude of the net force on π‘ŠοŠ§ per metre of length is 2.2Γ—10 N/m.

Key Points

  • Given the direction of a current in a straight wire, it is possible, using the right-hand rule, to determine the direction of the resulting magnetic field at any point in space.
  • The total magnetic field at any point in space equals the vector sum of all fields acting at that point.
  • Parallel, current-carrying wires exert forces on one another that are equal in magnitude.
  • The direction of these forces is discovered using the right-hand rule as follows: with the thumb of the right hand pointing in the direction of conventional current and the fingers in the direction of the net magnetic field, the palm faces the direction of the resulting force.
  • When two parallel wires carry current in the same direction, the forces between the wires attract. When the currents in the wire point in opposite directions, the resulting forces push the wires apart.

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