Lesson Explainer: The Magnetic Field due to a Current in a Circular Loop of Wire | Nagwa Lesson Explainer: The Magnetic Field due to a Current in a Circular Loop of Wire | Nagwa

Lesson Explainer: The Magnetic Field due to a Current in a Circular Loop of Wire Physics • Third Year of Secondary School

Join Nagwa Classes

Attend live Physics sessions on Nagwa Classes to learn more about this topic from an expert teacher!

In this explainer, we will learn how to calculate the magnetic field produced by a current in a circular loop of wire.

When a conducting wire has a current, it will produce a corresponding magnetic field like in the diagram below. The black lines show some of the field lines of the magnetic field.

This magnetic field depends on the direction of the current and the shape of the wire.

For example, let’s consider a curved wire. The magnetic field direction changes as the wire curves. The diagram below shows the magnetic field around three points along the wire, indicated by the green dots.

We can continue to bend this wire, making it into a semicircle. The magnetic field directions around the chosen points would then overlap at a specific point, as shown in the diagram below.

The further from the wire you go, the weaker the magnetic field strength is. However, if the wire is curved like this, the magnetic field lines can overlap and contribute to each other, producing a stronger field at this central point.

A wire curved to form a complete circle will produce a very strong magnetic field in one direction at its center.

The direction of this produced magnetic field can be determined by the right-hand screw rule. A right-handed screw can only be turned in one direction to drive it into a surface.

The direction we must turn the screw in is the direction of the current in a loop. The point of the screw, the direction it is being driven in, is the direction of the magnetic field at the center of that loop.

When a direction points into or out of the screen, we use the symbols below to show this.

These symbols are the same as what we would see if we looked at a screw that is pointing toward or away from us.

Let’s look back at the direction of the magnetic field from the diagram above.

The direction of the current in the loop is counterclockwise. Thus, the direction the screw must turn in is counterclockwise. The direction it is being driven in must be out of the screen. This means the magnetic field direction is also out of the screen.

Let’s look at an example.

Example 1: Direction of a Magnetic Field at the Center of a Coil

A circular loop of wire is carrying a constant current 𝐼 in a clockwise direction as viewed from above. The current creates a magnetic field. Based on the diagram, state the direction of the magnetic field at the center of the coil.

Answer

The magnetic field produced by a current-carrying wire loop will be in a single direction at the center. Using the right-hand screw rule, we know the direction the screw must turn is the same as the current in this loop: clockwise. Let’s look at this screw from a side angle of this loop.

In order for the screw to turn clockwise, it must be pointing into the loop. From the original perspective, this means the screw must be pointing into the screen.

Thus, the answer is D: into the screen.

Every point on the loop is contributing to the magnetic field strength at the center. For the strength of the field at the very center, there is a simple formula we can use.

Equation: Magnetic Field Strength at the Center of a Current-Carrying Wire Loop

The magnetic field strength 𝐡 at the center of a current-carrying wire loop is given by 𝐡=πœ‡πΌ2π‘Ÿ, where 𝐼 is the current in the wire loop, π‘Ÿ is the radius of the wire loop, and πœ‡οŠ¦ is the permeability of free space, often expressed as a value of 4πœ‹Γ—10 Tβ‹…m/A.

So, if the radius of a wire loop and its current are known, we can calculate its magnetic field strength.

As the current in a loop increases, so too does the magnetic field strength. Magnetic field strength is directly proportional to current.

As the radius increases, the magnetic field strength decreases. Magnetic field strength is inversely proportional to the loop’s radius. The diagram below shows two loops with the same current but different radii.

Since both loops have the same current, 𝐼, the only factor affecting the magnetic field strength is the radius. The loop with the larger radius, π‘ŸοŠ§, thus has a smaller magnetic field strength, 𝐡.

Let’s take a look at using this equation with some numbers. Suppose that we have a wire loop with a radius of 2.5 cm and a current of 1 A.

Before these values can be used to calculate the magnetic field strength at the center of this loop, we have to ensure the units match. The permeability of free space, πœ‡οŠ¦, is in tesla-metres per ampere. This means we want the units of length in metres, not centimetres.

To convert 2.5 cm, we know that there are 100 cm in 1 metre: 1100.mcm

Multiplying this by 2.5 cm gives 1100Γ—2.5=0.025.mcmcmm

We now have all the values we need to put into the wire loop magnetic field strength equation: 𝐡=πœ‡πΌ2π‘Ÿ.

The current is 1 A, π‘Ÿ is 0.025 m, and the permeability of free space is 4πœ‹Γ—10 Tβ‹…m/A: 𝐡=ο€Ή4πœ‹Γ—10β‹…/(1)2(0.025).TmAAm

The units of amperes cancel in the numerator, and the 2 is multiplied in the denominator: 𝐡=4πœ‹Γ—10β‹…0.05.Tmm

When dividing, the units of metres cancel, leaving behind only teslas: 4πœ‹Γ—10β‹…0.05=2.51Γ—10.TmmT

So, a wire loop with radius 2.5 cm and current 1 A has a magnetic field strength of 2.51Γ—10 T at its center.

Let’s look at an example question.

Example 2: Strength of a Magnetic Field at the Center of a Loop

A circular loop of wire carries a constant current of 0.9 A. The radius of the loop is 13 mm. Calculate the strength of the magnetic field at the center of the loop. Give your answer in teslas expressed in scientific notation to 1 decimal place. Use a value of 4πœ‹Γ—10 Tβ‹…m/A for πœ‡οŠ¦.

  1. 3.3Γ—10 T
  2. 1.4Γ—10 T
  3. 8.7Γ—10 T
  4. 3.5Γ—10 T
  5. 4.3Γ—10 T

Answer

The loop looks like this.

To find the magnetic field strength of this wire loop, we will use the equation 𝐡=πœ‡πΌ2π‘Ÿ.

We have the values of the current, 0.9 A, and the radius, 13 mm. Before directly using these values though, the units must match. We want the radius, 13 millimetres, to be in metres to match the permeability of free space.

There are 1β€Žβ€‰β€Ž000 mm in 1 m: 11000.mmm

Multiplying this by 13 mm gives 11000Γ—13=0.013.mmmmmm

So, the radius is 0.013 metres.

We can now put the values of 0.9 A, 0.013 m, and 4πœ‹Γ—10 Tβ‹…m/A into the equation: 𝐡=ο€Ή4πœ‹Γ—10β‹…/(0.9)2(0.013).TmAAm

The units of amperes cancel in the numerator as the current and permeability are multiplied together: 𝐡=3.6πœ‹Γ—10β‹…2(0.013).Tmm

The denominator is multiplied by 2 as follows: 𝐡=3.6πœ‹Γ—10β‹…0.026.Tmm

Dividing cancels the units of length, metres, leaving just teslas: 3.6πœ‹Γ—10β‹…0.026=4.349Γ—10.TmmT

So, the magnetic field strength at the center of this wire loop, rounded to 1 decimal place, is 4.3Γ—10 T.

The correct answer is E.

If the magnetic field strength at the center of the loop is known, then it can be used to find other variables in the equation.

As an example, let’s say that we have a loop of wire with an unknown radius. If the current in the wire and the magnetic field strength at the center is known, then the radius can be found. Starting with the base equation, we have 𝐡=πœ‡πΌ2π‘Ÿ.

In order to isolate the radius, we want π‘Ÿ on one side. We can accomplish this by multiplying both sides by π‘Ÿ as follows: π΅Γ—π‘Ÿ=πœ‡πΌ2π‘ŸΓ—π‘Ÿ.

This cancels the π‘Ÿ on the right side of the equation: π‘Ÿπ΅=πœ‡πΌ2.

To get π‘Ÿ by itself, we divide both sides by 𝐡: π‘Ÿπ΅π΅=πœ‡πΌ2𝐡.

The 𝐡s on the left side cancel, leaving π‘Ÿ=πœ‡πΌ2𝐡.

This new form of the equation can then be used to determine the radius of a loop.

Say that we have a loop like the one in the diagram below.

The radius is unknown, but the current is 1 A and the magnetic field strength at the center is 5Γ—10 T. To find the radius, we need to put the other given values into the new equation. Taking the permeability of free space as 4πœ‹Γ—10 Tβ‹…m/A, we have π‘Ÿ=πœ‡πΌ2π΅π‘Ÿ=ο€Ή4πœ‹Γ—10β‹…/(1)2(5Γ—10).TmAAT

The amperes in the numerator cancel as they are multiplied together: π‘Ÿ=4πœ‹Γ—10β‹…2(5Γ—10).TmT

The 2 is multiplied in the denominator: π‘Ÿ=4πœ‹Γ—10β‹…10Γ—10.TmT

When dividing these values, the teslas cancel, leaving behind only metres: π‘Ÿ=1.257Γ—10.m

Taking this value to 2 decimal places, the radius of the loop is 1.26Γ—10 metres.

The equation for the magnetic field strength at the center of a loop can also be rearranged for an unknown current. Looking again at the base equation: 𝐡=πœ‡πΌ2π‘Ÿ.

We begin isolating 𝐼 by multiplying both sides by 2π‘Ÿ as follows: 𝐡×2π‘Ÿ=πœ‡πΌ2π‘ŸΓ—2π‘Ÿ.

This cancels the 2π‘Ÿ on the right side of the equation: 2π‘Ÿπ΅=πœ‡πΌ.

We then divide both sides by the permeability of free space, πœ‡οŠ¦, as follows: 2π‘Ÿπ΅πœ‡=πœ‡πΌπœ‡.

This cancels the permeability of free space, leaving just the current: 2π‘Ÿπ΅πœ‡=𝐼.

Let’s look at an example that uses this form of the equation.

Example 3: Determination of the Current of a Wire in a Loop

A circular loop of wire with a radius of 9.5 cm carries a constant current of 𝐼 A. The strength of the magnetic field produced by the current is 5.2Γ—10 T at the center of the loop. Calculate 𝐼, rounding your answer to 1 decimal place. Use a value of 4πœ‹Γ—10 Tβ‹…m/A for πœ‡οŠ¦.

Answer

The current of this wire is unknown, but the magnetic field strength and radius are known. We can find the current using the modified equation for the magnetic field strength of a loop: 𝐼=2π‘Ÿπ΅πœ‡.

To get the correct units, we must first convert the radius from centimetres to metres. There are 100 cm in 1 m: 1100.mcm

Multiplying this by 9.5 cm gives 1100Γ—9.5=0.095.mcmcmm

The radius is 0.095 m and the magnetic field strength is 5.2Γ—10 T. The value we use for πœ‡οŠ¦ is 4πœ‹Γ—10 Tβ‹…m/A. So, we have 𝐼=2(0.095)ο€Ή5.2Γ—104πœ‹Γ—10β‹…/.mTTmA

Multiplying across the entire numerator gives units of tesla-metres (Tβ‹…m) on the top: 𝐼=9.88Γ—10β‹…4πœ‹Γ—10β‹…/.TmTmA

The division of these numbers causes the units of tesla-metres to cancel, leaving behind 1A in the denominator: 𝐼=9.88Γ—104πœ‹Γ—10.A

The 1A term in the denominator is equivalent to an A in the numerator. This is because dividing by a number is the same as multiplying by its reciprocal: 1=1Γ—1Γ—=.AAAA

So, when the numbers are divided, we are left with units of amperes: 9.88Γ—104πœ‹Γ—10=7.86.AA

Rounded to 1 decimal place, the current in the circular loop of wire is 7.9 amperes.

To increase the magnetic field strength at the center of a loop, the current can be increased or the radius can be decreased. There is another way to increase the magnetic field strength: by adding more loops.

When we have a set of loops that have the exact same radius and current, we determine the magnetic field strength at their center using the following equation.

Equation: Magnetic Field Strength at the Center of Multiple Current-Carrying Wire Loops

The magnetic field strength 𝐡 at the center of a series of wire loops, which have the same radius and carry the same current, is given by 𝐡=πœ‡π‘πΌ2π‘Ÿ, where 𝐼 is the current in the wire loops, π‘Ÿ is the radius of the wire loops, 𝑁 is the number of loops, and πœ‡οŠ¦ is the permeability of free space, often expressed as a value of 4πœ‹Γ—10 Tβ‹…m/A.

When 𝑁 is 1, the equation is the same as the base magnetic field strength at the center of a wire equation. Each additional loop multiplies the total magnetic field strength, so 2 loops doubles the strength, 3 triples it, and so on.

This equation is essentially just the single loop equation, but it is multiplied by the number of loops: 𝐡=π΅Γ—π‘πœ‡π‘πΌ2π‘Ÿ=πœ‡πΌ2π‘ŸΓ—π‘.multipleloopssingleloop

Let’s look back at the example of the loop with 1 A of current and a radius of 2.5 cm. If we were to line up 5 of these loops in a row, it would look like the diagram below.

Since there are 5 loops, the value of 𝑁 in the equation would be 5. The value of the current is still 1 A; the number of loops does not change this. We know from this previous example that the radius expressed in metres is 0.025 m. So putting these variables into the equation gives 𝐡=πœ‡π‘πΌ2π‘Ÿπ΅=ο€Ή4πœ‹Γ—10β‹…/(5)(1)2(0.025).TmAAm

Multiplying across the numerator cancels the units of amperes, leaving behind tesla-metres. The 5 loops do not contribute any units: 𝐡=6.28Γ—10β‹…2(0.025).Tmm

The 2 in the denominator is then multiplied across the radius: 𝐡=6.28Γ—10β‹…0.05.Tmm

Dividing these numbers cancels the metres, leaving behind the units of magnetic field strength, teslas: 6.28Γ—10β‹…0.05=1.26Γ—10.οŠͺTmmT

If we compare this value of 1.26Γ—10οŠͺ T to the value from a single loop of this radius and current, we would see it is 5 times as large.

Let’s look at an example.

Example 4: Magnetic Field Determination in Coil With Multiple Turns

A thin, circular coil of wire with a radius 4.2 cm carries a constant current of 3.9 A. The coil has 35 turns of wire. What is the strength of the magnetic field at the center of the coil? Give your answer in teslas expressed in scientific notation to 1 decimal place. Use πœ‡=4πœ‹Γ—10β‹…/TmA.

  1. 1.7Γ—10 T
  2. 4.1Γ—10 T
  3. 4.9Γ—10 T
  4. 5.8Γ—10 T
  5. 2.0Γ—10 T

Answer

Instead of perfectly lining up many loops, real-life electronics instead use a single thin wire that is tightly coiled together. In such cases, 𝑁 in the equation refers to the number of turns the wire has, rather than the number of individual loops.

This problem can be solved by using the equation 𝐡=πœ‡π‘πΌ2π‘Ÿ.

Before doing so, let’s convert the 4.2 centimetres into metres to match the units of πœ‡οŠ¦. There are 100 cm in 1 m: 1100.mcm

Multiplying this by 4.2 centimetres gives 1100Γ—4.2=0.042.mcmcmm

So, 4.2 cm is 0.042 m.

We can now put all the variables into the equation. The current is 3.9 A, the radius is 0.042 m, and the number of turns is 35: 𝐡=πœ‡π‘πΌ2π‘Ÿπ΅=ο€Ή4πœ‹Γ—10β‹…/(35)(3.9)2(0.042).TmAAm

The units of amperes cancel in the numerator as the terms are multiplied across: 𝐡=1.715Γ—10β‹…2(0.042).οŠͺTmm

Multiplying the 2 across the denominator gives 𝐡=1.715Γ—10β‹…0.084.οŠͺTmm

When dividing, the units of metres will cancel, leaving behind only teslas: 1.715Γ—10β‹…0.084=2.04Γ—10.οŠͺTmmT

So, to 1 decimal place, the magnetic field strength at the center of this series of turns in this current-carrying coil of wire is 2.0Γ—10 T, or E.

Let’s suppose now that we have a coil of wire where the magnetic field strength is known, but the number of turns is not.

To find the number of turns, we want to find the value of 𝑁. Looking at the equation 𝐡=πœ‡π‘πΌ2π‘Ÿ, we see that we need to isolate 𝑁. We can start doing this by multiplying both sides by 2π‘Ÿ: 𝐡×2π‘Ÿ=πœ‡π‘πΌ2π‘ŸΓ—2π‘Ÿ.

This cancels the 2π‘Ÿ on the right side: 2π‘Ÿπ΅=πœ‡π‘πΌ.

Now, we can divide both sides by πœ‡πΌοŠ¦2π‘Ÿπ΅πœ‡πΌ=πœ‡π‘πΌπœ‡πΌ.

This cancels the πœ‡πΌοŠ¦ term on the right side, leaving just 𝑁: 2π‘Ÿπ΅πœ‡πΌ=𝑁.

Let’s look at an example.

Example 5: Determination of the Number of Loops Using Magnetic Field Strength

A thin, circular coil of wire with a radius of 22 mm that has 𝑁 turns carries a constant current of 0.45 A. The strength of the magnetic field produced by the current is 2.3Γ—10οŠͺ T at the center of the coil. Calculate 𝑁 to the nearest whole number of turns. Use a value of 4πœ‹Γ—10 Tβ‹…m/A for πœ‡οŠ¦.

Answer

The modified form of the equation that can be used to find 𝑁 is 𝑁=2π‘Ÿπ΅πœ‡πΌ.

Before we can use this equation, we need to ensure all the variables have matching units. This means the radius of 22 mm needs to be converted to metres.

There are 1β€Žβ€‰β€Ž000 millimetres in 1 metre: 11000.mmm

Multiplying this by 22 mm gives 11000Γ—22=0.022.mmmmmm

So, 22 mm is 0.022 metres.

We now put the values into the equation for 𝑁. The current is 0.45 A, the radius is 0.022 m, the magnetic field strength is 2.3Γ—10οŠͺ T, and πœ‡οŠ¦ is 4πœ‹Γ—10 Tβ‹…m/A. This gives us 𝑁=2π‘Ÿπ΅πœ‡πΌπ‘=2(0.022)ο€Ή2.3Γ—10(4πœ‹Γ—10β‹…/)(0.45).οŠͺmTTmAA

Multiplying across the numerator gives units of tesla-metres as follows: 𝑁=1.012Γ—10β‹…(4πœ‹Γ—10β‹…/)(0.45).TmTmAA

Multiplying across the denominator cancels the units of amperes, leaving behind units of tesla-metres: 𝑁=1.012Γ—10β‹…5.65Γ—10β‹….TmTm

When dividing these two numbers, the units cancel completely. This is ideal, since the number of turns is dimensionless: 1.012Γ—10β‹…5.65Γ—10β‹…=17.89.TmTm

Rounded to the nearest whole number, this coil of wire thus has 18 turns.

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

Key Points

  • The direction of a magnetic field at the center of a loop of wire is given by the right-hand screw rule.
  • The magnetic field strength 𝐡 at the center of a current-carrying loop of wire is given by the equation 𝐡=πœ‡πΌ2π‘Ÿ, where 𝐼 is the current in the loop, π‘Ÿ is the radius is the loop, and πœ‡οŠ¦ is the permeability of free space, often expressed as 4πœ‹Γ—10 Tβ‹…m/A.
  • When there are multiple current-carrying loops of the same radius in a coil, the magnetic field strength 𝐡 is given by the equation 𝐡=πœ‡π‘πΌ2π‘Ÿ, where 𝐼 is the current in the loop, π‘Ÿ is the radius is the loop, 𝑁 is the number of turns or loops in the wire, and πœ‡οŠ¦ is the permeability of free space, often expressed as 4πœ‹Γ—10 Tβ‹…m/A.

Join Nagwa Classes

Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!

  • Interactive Sessions
  • Chat & Messaging
  • Realistic Exam Questions

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy