Magnetic Field of a Current-Carrying Wire

Have you found yourself interested in how electric and magnetic fields are linked? The laws of electromagnetism state that electric charges and magnetic fields are closely intertwined, with the movement of charged particles generating a surrounding magnetic field. One example of this is current-carrying wires; as currents are the movement of electrons, which are negatively charged particles, their movement throughout the wire generates a magnetic field around the wire. Keep reading to learn more about these fields and how they interact with other wires!

Create learning materials about Magnetic Field of a Current-Carrying Wire with our free learning app!

• Flashcards, notes, mock-exams and more
• Everything you need to ace your exams

Magnetic Field of a Current-Carrying Wire Formula

Firstly, let's define the equation that allows us to calculate the magnetic field generated by a current-carrying wire. It is given as

$B = \frac{\mu_{0}}{2\pi}\frac{I}{r},$

where $$B$$ is the magnitude of the magnetic field measured in teslas $$\mathrm{T}$$, $$\mu_{0}$$ is the permeability of free space given by a value of $$4\pi \times 10^{-7} \,\mathrm{\frac{H}{m}}$$ where $$\mathrm{H}$$ denotes henrys,

$$I$$ is the current in the wire measured in amperes $$\mathrm{A}$$, and $$r$$ is the radial distance away from the wire measured in meters $$\mathrm{m}$$.

Let's analyze this equation more; the magnitude of the magnetic field generated by a straight current-carrying wire is proportional to the current in the wire. Thus, by running a larger current through the wire, we achieve a stronger magnetic field. On the other hand, the magnitude of the magnetic field is inversely proportional to the radial distance from the wire. Therefore, the further away we are, the weaker the magnetic field is.

Fig. 1 - The magnetic field generated by a straight current-carrying wire.

From the figure above, the magnetic field is denoted by the pink circles, highlighting that the generated field is tangent to the current-carrying wire and concentric circles with their center as the wire. As the field lines are tangent to the wire, there is no component of the magnetic field toward the wire.

Direction of Magnetic Field around a Current-Carrying Wire

So how can we determine the direction of the generated magnetic field? We can use something called the right-hand grip rule. Referring to the figure below, we orient your right hand with the thumb pointing upwards whilst the fingers are curved towards your palm. In this case, your thumb is pointed in the direction of the current whilst the fingers are curved in the direction of the magnetic field.

Let's consider an example.

Fig. 3 - The direction of the magnetic field in a current-carrying wire is determined by the right-hand grip rule.

This is the wire we saw in the previous section; by using the right-hand grip rule, we can point our thumb in the direction of the current in the wire, ie. pointing towards the right. By orienting our hands in this direction, our fingers naturally curl in an anti-clockwise direction, which is the resultant direction of the magnetic field, denoted by the arrows on the magnetic field lines.

Magnetic Field Strength of a Current-Carrying Wire

Now let's see how the magnetic field strength of a current-carrying wire affects other objects, for instance, another wire.

Magnetic Field Strength between Two Current-Carrying Wires

Now we have established how to calculate the magnetic field generated by the current in a wire, let's look at how these magnetic fields may interact with one another. If you have read our other article on the magnetic field of a moving charge, you may be familiar with how moving charges interact with magnetic fields via the Lorentz force. If not, that's okay, here's a quick refresher!

When charged particles travel through a magnetic field, they experience something called the Lorentz force, which is given by

$\vec{F}_{\text{M}} = q \vec{v} \times \vec{B},$

where $$\vec{F}_{\text{M}}$$ is the Lorentz force measured in newtons $$\mathrm{N}$$, $$q$$ is the charge of the particle measured in coulombs $$\mathrm{C}$$, $$\vec{v}$$ is the velocity of the charged particle measured in $$\mathrm{\frac{m}{s}}$$, and $$\vec{B}$$ is the magnetic field measured in Teslas $$\mathrm{T}$$. Furthermore, we can also represent the magnitude of the Lorentz force as

$|\vec{F}_{\text{M}}| = |q\vec{v}| |\sin(\theta)| |\vec{B}| ,$

where $$\theta$$ is the angle between the trajectory of the charged particle and the magnetic field $$B$$, and all the other quantities are the same as above.

The magnitude of the Lorentz force on a wire is given as

$|\vec{F}_{\text{M}}| = |I \vec{L}| |\sin(\theta) | |\vec{B} | ,$

where $$I$$ is the current running in the wire measured in $$\mathrm{A}$$, $$L$$ is the length of the wire in the magnetic field measured in $$\mathrm{m}$$, and all the other quantities are the same as before. The direction of the resultant Lorentz force can be determined using the left-hand rule.

Derivation of Lorentz Force on a Wire

This derivation is a deep dive and is not needed in the scope of the course.

Firstly, let's rewrite the current in terms of charge, this is given as

$I = \frac{q}{t},$

where $$q$$ is the total charge of the electrons measured in coulombs $$\mathrm{C}$$, $$I$$ is the current measured in amperes $$\mathrm{A}$$, and $$t$$ is the total time the charges took to travel measured in seconds $$\mathrm{s}$$.

Furthermore, we can rewrite the velocity $$v$$ of the charged particles as

$|\vec{v}| = \frac{L}{t},$

where $$L$$ is the distance traveled by the particles measured in $$\mathrm{m}$$, and $$t$$ is the time taken measured in $$\mathrm{s}$$. We can substitute the velocity $$v$$ into our magnitude of the Lorentz force to find that

\begin{align} |\vec{F}_{\text{M}}| &= |\vec{B}|q |\vec{v}| |\sin(\theta)| \\ |\vec{F}_{\text{M}}| &= |\vec{B}| q \frac{L}{t} |\sin(\theta)|. \end{align}

Going even further, we can substitute the time $$t$$ from our expression of current to find

\begin{align} |\vec{F}_{\text{M}}| &= |\vec{B}| q \left(\frac{L}{\frac{q}{I}} \right) |\sin(\theta)| \\ |\vec{F}_{\text{M}}| &= |\vec{B}| I L |\sin(\theta)|, \end{align}

which is the expression we established before.

Right-Hand Rule to find the Magnetic Field of a Current-Carrying Wire

Now let's consider two wires running parallel to one another as in the figure below.

Fig. 4 - Two parallel wires exerting a force on one another.

Consider the force $$F_1$$ acting on the wire on the right-hand side carrying current $$I_2$$. The magnetic field generated from the wire on the left is acting downwards on the wire on the right, whilst the current is running into the page. If we use our left-hand rule and orient the first finger downwards, whilst the second finger points into the page, we find that the resultant force $$F_1$$ (given by the direction of the thumb) is pointing towards the left.

Carrying out a similar process with the wire on the left, this time instead of the magnetic field from the wire on the right acting upwards, we find that the resultant Lorentz force $$F_2$$ is pointing towards the right. Thus we can conclude that two wires carrying currents in the same direction will experience an attractive force toward one another.

Can you find the direction of the force if the currents are running in opposite directions? Look to our flashcards for this article to test yourself!

Formula of the Magnetic Force between Two Current-Carrying Wires

Consider again the situation above where the two wires are carrying currents in the same direction. Using the magnitude of the Lorentz force equation, we can substitute our expression for the magnetic field from wire 1, to find the force on wire 2 as

\begin{align} F &= BI_2 L\sin(\theta) \\ F &= \frac{\mu_0}{2\pi}\frac{I_1}{r} I_2 L \sin(90^{\circ}) \\ F &= \frac{\mu_0}{2\pi} \frac{I_1 I_2}{r} L . \end{align}

Here the first line of our derivation indicates the force on the second wire, which is why we have replaced the $$I$$ term with $$I_2$$. Then, we substitute the expression of the magnetic field from the first wire, which results in the inclusion of the $$I_1$$ term.

Eliminating the trigonometry term as that goes to one, we can rearrange to find the force per unit length on the second wire as

$\frac{F}{L} = \frac{\mu_0}{2\pi} \frac{I_1 I_2}{r} ,$

where $$\frac{F}{L}$$ is the force per unit meter measured in newtons per meter $$\mathrm{N}{m}$$, $$r$$ is the radial distance from the second wire, and all the other terms are the same as previous equations.

Magnetic Field due to Current-Carrying Circular Loop

We now look at a specific case of the shape of a current-carrying wire; in the case of the wire forming a loop, our formula for the magnetic field strength around the wire changes slightly. In this instance, our equation becomes

$B = \frac{\mu_0 I}{2R} ,$

where $$B$$ is the magnitude of the magnetic field strength at the center of the loop measured in teslas $$\mathrm{T}$$, $$R$$ is the radius of the current loop, and all the other quantities remain the same as our previous equation.

Thus, we can see that the formula looks very similar to the general magnetic field equation around a wire, but some small differences need to be memorized. For instance, we no longer have the factor of $$2\pi$$ on the denominator as the shape of the surrounding magnetic field is no longer in circular rings. Additionally, the radius $$R$$ is no longer a variable but rather a constant.

Magnetic Force on Current-Carrying Wires Example

Finally, let's consider an example question.

We have a wire of length $$1.5\,\mathrm{m}$$, carrying a current of $$2.7\,\mathrm{mA}$$ in the upwards direction.

1. What is the magnitude of the magnetic field generated by the wire at a radial distance of $$6.7\,\mathrm{cm}$$ away?

Now we have another wire of length $$1.2\,\mathrm{m}$$, carrying a current of $$9.7\,\mathrm{mA}$$ in the downwards direction. It is placed at a radial distance of $$6.7\,\mathrm{cm}$$ away from the first wire.

1. What is the direction of the Lorentz force acting on the second wire?
2. What is the magnitude of this force?

The diagram below shows the setup of the question.

Fig. 5 - Two wires carrying currents in opposite directions.

1. To calculate the magnitude of the magnetic field, let's use our equation for the magnetic field from a current running in a wire. Substituting the numbers from the question we find

$B = \frac{4\pi \times 10^{-7} \,\mathrm{\frac{H}{m}} \times 2.7 \times 10^{-3}\,\mathrm{A}}{2\pi \times 6.7 \times 10^{-2} \,\mathrm{m}} = 8.1\times 10^{-9}\,\mathrm{T} ,$

where we have used the fact that $$1 \, \mathrm{H} = 1 \, \mathrm{\frac{ kg \, m^2}{s^2 \, A^2}}$$ and $$1\, \mathrm{T} = 1 \, \mathrm{\frac{kg}{s^2 \, A}}$$.

2. To find the direction of the Lorentz force on the second wire, we can use both the right-hand grip rule and the left-hand rule. Firstly, we need to determine the direction of the magnetic field coming from wire one. Pointing our right thumb in the direction of the current reveals that the field is moving in a clockwise orientation.

Subsequently, considering the orientation of the two wires, it can then be deduced that the direction of the magnetic field is acting downwards onto wire 2. Therefore, pointing our first finger downwards and our second finger in the direction of the current shows that the resultant Lorentz force is to the right.

If we determined the direction of the Lorentz force on wire 1, we would find that the resultant force would be to the left. Therefore, when two wires parallel to one another have currents running in opposite directions, they experience a repulsive force.

3. Finally, we need to calculate the magnitude of the force acting on the second wire. Using our equation before, we can substitute our numbers to find

\begin{align} F &= 8.1 \times 10^{-9} \, \mathrm{T} \times 9.7 \times 10^{-3} \, \mathrm{A} \times 1.2 \, \mathrm{m} \times \sin \left(\frac{\pi}{2} \right) \\ &= 9.4 \times 10^{-11} \, \mathrm{N}, \end{align}

where we have used the fact that $$1 \, \mathrm{N} = 1 \, \mathrm{\frac{kg \, m}{s^2}}$$ and $$\theta = \frac{\pi}{2} \, \mathrm{rads}$$ because the direction of the magnetic field and the wire are perpendiculars.

Magnetic Field of a Current-Carrying Wire - Key takeaways

• A current running through a wire generates a magnetic field around it.
• The magnetic field is in the shape of concentric circles with their center at the wire.
• The magnitude of the magnetic field is given by $$B = \frac{\mu_0}{2\pi}\frac{I}{r}$$.
• The direction of the magnetic field can be determined by the right-hand grip rule.
• Two current-carrying wires, placed next to one another, will exert a Lorentz force unto one another, given by $$F = BIL \sin(\theta)$$.
• If the currents are running in the same direction, they experience an attractive force.
• If the currents are running in the opposite direction, they experience a repulsive force.

References

1. Fig. 1 - Magnetic field coming out of current-carrying wire, StudySmarter Originals.
2. Fig. 2 - Right-hand grip rule, StudySmarter Originals.
3. Fig. 3 - Direction of the magnetic field due to current-carrying wire, StudySmarter Originals.
4. Fig. 4 - Two parallel wires exerting a force on one another, StudySmarter Originals.
5. Fig. 5 - Two parallel wires carrying currents in opposite directions, StudySmarter Originals.

Flashcards inMagnetic Field of a Current-Carrying Wire 15

Learn with 15 Magnetic Field of a Current-Carrying Wire flashcards in the free StudySmarter app

We have 14,000 flashcards about Dynamic Landscapes.

What is the magnetic field inside a current-carrying wire?

The magnetic field inside a current-carrying wire can be found using Ampere's law as $$B = \frac{\mu r I}{2 \pi R^2}$$.

How do you calculate the magnetic field from a current-carrying wire?

We can calculate the magnetic field from a current-carrying wire as $$B = \frac{mu_0}{2\pi} \frac{I}{r}$$.

Is the magnetic field zero inside a conductor?

No, it can be calculated using Ampere's law.

What is the magnetic field a distance d from a wire?

It can be calculated using $$B = \frac{\mu_0}{2\pi} \frac{I}{d}$$, where $$d$$ is the radial distance from the wire.

What is the strength of a magnetic field called?

$$B$$ and is measured in teslas $$\mathrm{T}$$.

Is there a magnetic field inside a hollow conductor?

No, there will be no current enclosed so by Ampere's law, there is no magnetic field.

What is the unit of a magnetic field?

Teslas $$\mathrm{T}$$.

Test your knowledge with multiple choice flashcards

What is the expression for the magnetic field generated by a current-carrying wire?

What is the shape of the magnetic field generated by a current-carrying wire?

How does the strength of the magnetic field vary with radial distance $$r$$ from the wire?

StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

StudySmarter Editorial Team

Team Magnetic Field of a Current-Carrying Wire Teachers

• Checked by StudySmarter Editorial Team