Lesson Explainer: Force on Conducting Wires in Magnetic Fields Physics • 9th Grade

In this explainer, we will learn how to use the formula 𝐹=𝐵𝐼𝐿 to calculate the force experienced by a current-carrying wire that has been placed in a uniform magnetic field.

You have probably seen how permanent magnets interact with each other before, repelling or attracting each other depending on which poles are facing. They do this by using magnetic fields, abbreviated as 𝐵, the direction of which would look like the diagram below for opposite poles.

Magnets do not only exert force on other magnets, however. In the right conditions, they can also exert force on a wire. Let’s say that we put a wire in between these magnets such that it is within the magnetic field like in the diagram below.

For a wire with no current, there is no force, but if this wire has a current, 𝐼, perpendicular to the magnetic field direction, then the magnetic field interacts with it, creating a force on the wire out of the screen.

This force may potentially cause the wire to move, just like permanent magnets may move when exposed to each other’s magnetic fields. The force on this wire is shown in the diagram below in yellow.

The dot with the circle around it indicates a force going out of the screen, while a circle with an X means into the screen, as demonstrated in the figure below.

The exact direction of the force can be found using Fleming’s left-hand rule. Using your left hand, extend your index (pointer) finger forward, point your thumb upward, and turn your middle finger perpendicular to your index finger, as shown in the figure below.

Your index finger is the direction of the magnetic field, 𝐵, your middle finger the direction of current, 𝐼, and your thumb is the direction of the force exerted on the wire, 𝐹.

Let’s look at an example.

Example 1: Understanding the Direction of the Force Experienced by a Current-Carrying Wire in a Uniform Magnetic Field

The diagram shows a section of wire that has been positioned at 90 to a 0.1 T magnetic field. The wire carries a current of 2 A. What is the direction of the force acting on the wire due to the magnetic field?

Answer

Though we are given the values of magnitude and current, this has no bearing on the direction of current. To find the direction of the force that acts on the wire when the wire is at a 90 angle, we use the left-hand rule.

The direction of the magnetic field, the index finger, is to the right. The direction of current, the middle finger, is upward. The palm of the hand should be facing upward.

Sticking out the thumb, we see that the the direction of the force acting on the wire will be going into the screen, so we would indicate it using a circle with an X through it.

The direction of the force acting on the wire is into the screen.

The closer to the full 90 degrees the directions of the magnetic field and current are, the stronger the force on the wire is. The value of the force gradually goes down until the force is 0, when the directions are parallel, or at 0 degrees, as seen in the diagram below.

The direction of the current in this case would not matter, since it is parallel either way.

Let’s look at an example.

Example 2: Understanding the Effect of a Uniform Magnetic Field on a Current-Carrying Wire That Is Parallel to the Field

The diagram shows a section of wire that has been positioned parallel to a uniform 0.1 T magnetic field. The wire carries a current of 2 A. What is the direction of the force acting on the wire due to the magnetic field?

Answer

The direction of force does not depend on the values of magnetic field strength or current, even though these values are given. What matters here is the angle.

Only where there is a nonzero angle between the direction of current and direction of the magnetic field is there a force acting on the wire.

The magnetic field and direction of current in the wire are parallel in this example, so there is no force acting on the wire in any direction.

The way we can calculate the force acting on these wires is by using the following equation.

Equation: Force on a Conducting Wire in a Magnetic Field

When a wire with a current has a direction perpendicular to a magnetic field’s direction, the magnetic field acts on the wire with force 𝐹: 𝐹=𝐵𝐼𝐿, where 𝐵 is the strength of the magnetic field, 𝐼 is the current in the wire, and 𝐿 is the length of wire that is passing through the magnetic field.

Let’s look at an example using this equation.

Example 3: Finding the Force Acting on a Current-Carrying Wire in a Uniform Magnetic Field

A 20 cm section of wire carrying a current of 12 A is positioned at 90 to a 0.1 T magnetic field. What is the size of the force acting on the wire?

Answer

We can start by looking at the variables we need to find in the equation that relates the force on a current-carrying wire: 𝐹=𝐵𝐼𝐿.

𝐵 is the magnetic field strength, which we are given in its SI unit, tesla (T). Expressed another way, 1 tesla is equal to 1 newton per ampere-metre, NAm×. This magnetic field has a strength of 0.1 T.

The current, 𝐼, is given as 12 amperes.

We are looking at a 20 cm section of wire, so this is the value of 𝐿. We want it in terms of regular metres, though, in order to cancel with the metres in teslas. There are 100 cm in 1 metre, which expressed as a relation looks like 1100.mcm

So, multiplying this relation with the given value of 20 cm puts it in terms of metres: 1100×20=0.2.mcmcmm

This wire is 0.2 metres long.

We have all the variables we need to find the force acting on the wire. Substituting in the magnetic field strength, 0.1 T, current, 12 A, and length, 0.2 m, gives 𝐹=𝐵𝐼𝐿𝐹=(0.1)(12)(0.2).TAm

Now, let’s expand out the units of tesla to see how they cancel with other units when multiplied together: 𝐹=0.1×(12)(0.2).NAmAm

We can now see that the metres and amperes cancel when these terms are multiplied together, leaving behind only newtons, the SI units of force: 0.1×(12)(0.2)=0.24.NAmAmN

So, the size of the force acting on the wire is equal to 0.24 newtons.

If needed, we can also isolate specific variables in the equation that we wish to find, provided we are given the others.

Let’s say that we have a wire with a current of 5 A with a direction perpendicular to a 1 T magnetic field strength. The force on the wire is 0.4 N, but we do not know the length of the wire through the magnetic field, 𝐿. Such a wire is shown in the figure below.

We can find the length by looking at the equation for force on a current-carrying wire through a magnetic field, then isolating 𝐿 on both sides: 𝐹=𝐵𝐼𝐿.

We begin doing this by dividing both sides by 𝐵𝐼: 𝐹𝐵𝐼=𝐵𝐼𝐿𝐵𝐼.

This cancels the 𝐵𝐼 on the right side, leaving behind only the length, 𝐿: 𝐹𝐵𝐼=𝐿.

Substituting in the other values, we can solve for the length: (0.4)(1)(5)=𝐿.NTA

We expand out the units of tesla first: (0.4)1(5)=𝐿.NANAm×

Dividing by a number is the same as multiplying but its reciprocal. When the units of newtons and amperes cancel, it will look something like this: NANAAmNm=××=.NAm×

This means all the units except length cancel, giving an answer of (0.4)1(5)=0.08.NAmNAm×

The length of this wire is a mere 0.08 metres, or 8 centimetres.

The same process can be done for finding the current. Say we have a wire with an unknown value of current, but we know that its current direction is perpendicular to a magnetic field with a strength of 1 T. The force on the wire is 0.4 N, and the length of the wire passing through the magnetic field is 2 m. Such a wire is shown in the figure below.

Let’s find the current. Beginning with the original equation 𝐹=𝐵𝐼𝐿, we can divide both sides by 𝐵𝐿: 𝐹𝐵𝐿=𝐵𝐼𝐿𝐵𝐿.

This cancels the 𝐼𝐿 on the right, leaving only the current, 𝐼: 𝐹𝐵𝐿=𝐼.

Let’s then substitute in the values of the force, 0.4 N, magnetic field strength, 1 T, and length, 2 m, to get (0.4)(1)(2)=𝐼NTm and then expand out the units of tesla to get (0.4)1(2)=𝐼.NmNAm×

Dividing by a number is the same as multiplying but its reciprocal. When the units of newtons and metres cancel, it will look something like this: NmNmAmNA=××=.NAm×

This means all the units but current cancel, leaving only amperes: (0.4)1(2)=0.2.NmANAm×

So, the current in this wire is 0.2 A.

Let’s look at an example for finding magnetic field strength.

Example 4: Finding the Strength of a Uniform Magnetic Field from the Force Experienced by a Current-Carrying Wire

When positioned at 90 to a magnetic field, a wire of length 1 m carrying a current of 4 A experiences a force of 0.2 N. What is the strength of the magnetic field?

Answer

This wire is positioned at 90 degrees to the magnetic field, meaning it is fully perpendicular to the magnetic field direction.

We want to isolate the magnetic field strength, 𝐵. Starting with the base equation 𝐹=𝐵𝐼𝐿, we divide both sides by 𝐼𝐿: 𝐹𝐼𝐿=𝐵𝐼𝐿𝐼𝐿.

This causes 𝐼𝐿 on the right side to cancel, leaving just 𝐵: 𝐹𝐼𝐿=𝐵.

Let’s then substitute the values of the force, 0.2 N, current, 4 A, and length, 1 m, to get (0.2)(4)(1)=𝐵.NAm

The units of tesla are newtons per ampere-metre, which is exactly what we see here. Multiplying them together gives the answer of (0.2)(4)(1)=0.05.NAmT

The strength of the magnetic field is 0.05 teslas.

The force that is acting on a wire from a magnetic field can cause the wire to move, but the form is not always as simple as a long straight wire.

Let’s look at an example.

Example 5: Understanding the Forces Experienced by a Square of Current-Carrying Wire in a Uniform Magnetic Field

The diagram shows a square section of a wire that has been positioned in a uniform magnetic field such that two of its sides are perpendicular to the direction of the field and the other two sides are parallel to the field. The magnetic field has a strength of 0.3 T, and the current through the wire is 2 A. Each side of the square is 0.2 m long.

  1. What is the magnitude of the force acting on the right-hand side of the square?
    Initially, what is the direction of the force acting on the right-hand side of the square?
    1. Out of the screen
    2. Into the screen
  2. What is the magnitude of the force acting on the left-hand side of the square?
    Initially, what is the direction of the force acting on the left-hand side of the square?
    1. Into the screen
    2. Out of the screen
  3. What is the magnitude of the force acting on the top side of the square?
    What is the overall effect of the magnetic field on the wire?
    1. The magnetic field has no overall effect on the wire.
    2. The magnetic field makes the wire spin about the 𝑦-axis of the screen.
    3. The magnetic field accelerates the wire into the screen.
    4. The magnetic field accelerates the wire out of the screen.
    5. The magnetic field makes the wire spin about the 𝑥-axis of the screen.

Answer

Part 1

The direction of the current in the wire is perpendicular to the magnetic field, so it has a force acting on it. We can find this force using the equation 𝐹=𝐵𝐼𝐿, then substituting in the values of the magnetic field strength, 0.3 T, current, 2 A, and length, 0.2 m, gives 𝐹=(0.3)(2)(0.2).TAm

Expanding out the units of tesla shows that amperes and metres cancel to give 0.3×(2)(0.2)=0.12.NAmAmN

So, the wire has a force of 0.12 newtons.

The direction the force is pointed in can be found using Fleming’s left-hand rule. The magnetic field (index finger) points to the right, and the current (middle finger) points straight up. This means that extending the thumb, with the palm facing upward, makes the force point inward toward the screen.

The correct answer is B: into the screen.

Part 2

Just like in the first part, the direction of the current in the wire is perpendicular to the magnetic field, only it is in the opposite direction. We can find the magnitude of the force using the equation 𝐹=𝐵𝐼𝐿, then substituting in the given values of the magnetic field strength, 0.3 T, current, 2 A, and length, 0.2 m, gives 𝐹=(0.3)(2)(0.2).TAm

Expanding out the units of tesla shows that amperes and metres cancel to give 0.3×(2)(0.2)=0.12.NAmAmN

The magnitude of the force is the same as in part 1, 0.12 newtons.

The direction of the force, though, is different. The magnetic field points to the right, and the current is pointing down. This makes our thumb stick out away from the screen, with the palm facing down.

The correct answer is B: out of the screen.

Part 3

Along the top of the square, the current is parallel to the magnetic field direction. It does not matter what the direction of the current is, as two lines can be parallel even when traveling in opposite directions. The magnitude of the force acting on the top of the square is thus 0.

The same is true for the bottom of the square: there is no force on the wire there.

Knowing all of this, let’s think of how all the forces are acting on this loop of wire. The top and bottom parts have no force, while the right and left portions of the wire have constant force pointing into and out of the wire respectively. Since these forces do not oppose one another, there is an overall effect on the wire. This is not the case in A.

The wire on the sides are not adding together either, since they are pointing in opposite directions. This means that the loop will not accelerate into or out of the screen, so the answer cannot be C or D.

Rather, the wire will constantly rotate around the center. Since the top and bottom have no force, it will not rotate at all along the 𝑥-axis. Looking at it from a slightly different angle, it would appear as shown in the diagram.

It will begin to rotate along the 𝑦-axis.

So, the correct answer is B: it will spin around the 𝑦-axis.

We have seen that the force on a conducting wire in a magnetic field is 0 when the wire is parallel to the magnetic field direction, but when it is perpendicular, the force can be found using the equation 𝐹=𝐵𝐼𝐿.

These two cases are next to each other on the diagrams below.

If, however, a conducting wire is at a different angle other than 0 or 90 degrees, it must be expressed with a different equation. A wire at a different angle is shown in the diagram below, its angle being 𝜃 degrees.

Such a wire will have a force acting modified by a cosine(𝜃), as described in the equation below.

Equation: Force on a Conducting Wire at an Angle in a Magnetic Field

When a wire with a current has a direction that makes an angle to a magnetic field direction, the magnetic field acts on the wire with force 𝐹: 𝐹=𝐵𝐼𝐿(𝜃),cos where 𝐵 is the strength of the magnetic field, 𝐼 is the current in the wire, 𝜃 is the angle the wire makes with the magnetic field, and 𝐿 is the length of wire that is passing through the magnetic field.

The diagram below shows all of these variables outlined together.

Let’s summarize what we have learned in this explainer.

Key Points

  • A current-carrying wire placed in a magnetic field may experience a force.
  • If the direction of the magnetic field is perpendicular (90) to the current in a wire, the force the wire experiences due to the magnetic field 𝐹 is 𝐹=𝐵𝐼𝐿, where 𝐵 is the magnetic field strength, 𝐼 is the magnitude of the current in the wire, and 𝐿 is the length of the wire that is in the field.
  • If the direction of the magnetic field is parallel (0) or antiparallel (180) to the current in a wire, the wire experiences no force: 𝐹=0.
  • Use Fleming’s left-hand rule to determine the direction of the force from the magnetic field: index finger pointed forward for magnetic field direction, middle finger for current direction, and the thumb pointed upward for force from the magnetic field.

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