Physics Lesson 16.2.3 - Magnetic Field Produced by Electric Currents. Right Hand Rule

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Welcome to our Physics lesson on Magnetic Field Produced by Electric Currents. Right Hand Rule, this is the third lesson of our suite of physics lessons covering the topic of Magnetic Field Produced by Electric Currents, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

Magnetic Field Produced by Electric Currents. Right Hand Rule

In this part of the tutorial, we will discuss about the magnetic field produced through a number of methods. They vary from each other in formulae but have one thing in common: the direction of magnetic field is found using the so-called "right-hand rule." It consists on grasping the wire by the right hand where the thumb lies in the direction of current flow. The other four fingers show the direction of magnetic field as shown in the figure.

Physics Tutorials: This image provides visual information for the physics tutorial Magnetic Field Produced by Electric Currents

a. Magnetic Field Produced by a Long Straight Wire

When introducing the Oersted experiment, we pointed out the fact that the needle of compass deflects from its normal direction when placed near a current carrying wire. The direction of magnetic field produced by a long current wire is found using the right-hand rule discussed in the previous paragraph.

In addition, when introducing the concept of magnetic induction (which is a quantity related to the magnetic field strength), we explained that magnetic field is inversely proportional to the distance r from the current carrying wire and directly proportional to the magnitude of current I. This joint proportionality is written as

B⃗ ∝ l/r

A proportion becomes equality when multiplying the expression by a constant. Since the magnetic field around a long current carrying wire is circular, we obtain equal values for magnetic induction is constant in the entire circumference of the circle of radius r around the wire.

In addition, during the calculations we must consider a quantity similar to the vacuum permittivity ϵ0 which is used in formulae when dealing with electric field. This quantity is a constant known as the magnetic permeability of free space (vacuum), μ0. It has a value of

μ₀ = 4π × 10^-7 N/A²

Thus, when considering all the above factors, the formula of magnetic induction (in scalar form) produced by a long current carrying wire is

B = μ₀ ∙ I/2π ∙ r

Example 2

What is the magnetic induction at a distance of 40 cm from a long conducting wire carrying a 0.3 A current?
What is the direction of magnetic field if the current Is flowing vertically down?

Solution 2

Clues:

r = 40 cm = 0.40 m
I = 0.3 A
μ0 = 4π × 10-7 N/A2
a) B = ?
b) Direction of B = ?

From the equation of magnetic induction in a long current carrying conducting wire, we have:
B = μ₀ ∙ I/2π ∙ r
= 4π × 10^-7 N/A² ∙ 0.3 A/2π ∙ 0.40 m
= 0.75 × 10^-7 N/A ∙ m
= 7.5 × 10^-8 T
When applying the right hand rule as shown in the figure below, we find the direction of magnetic induction shown by the four fingers as shown in the figure. We say the direction of magnetic induction is clockwise when looked from up. Since it is impossible to plot a 3-D figure every time we have to solve such exercises, we use a simpler notation to represent the direction of current when the situation is viewed from up. Thus, when the current enters the paper we use a x-symbol inside a circle while when the current Is flowing out of paper, we use a dot inside a circle to represent the direction of current. The magnetic field is easier when using such notation, as we have only to write concentric circles to represent it. Look at the figure: The above symbols are also used to represent the direction of magnetic field (induction) direction when a 3-D figure must be expressed in two dimensions only. Look at the figure.

Example 3

Two long parallel wires carrying currents of I and 3I respectively, have a distance d from each other. Calculate the resultant magnetic field at any midpoint position if:

The currents have the same direction
The currents have opposite direction

Solution 3

When two parallel wires have currents flowing in the same direction, they will produce magnetic fields in opposite direction at any midpoint d/2 between them. Look at the figure below: The magnetic induction B₁ at the distance d/2 from the wire I is
B₁ = μ₀ ∙ I/2π ∙ d/2
= μ₀ ∙ I/π ∙ d
The magnetic induction B₂ at the distance - d/2 from the wire 3I is
B₂ = μ₀ ∙ 3I/2π ∙ (- d/2)
= -3μ₀ ∙ I/π ∙ d
= -3B₁
(The negative sign is because the second wire is on the other side of midpoint to the first wire.)
Therefore, the net magnetic induction at the midpoint of the two wires is
B = B₁ + B₂
= B₁ - 3B₁
= -2B₁
(Or we can say "the net magnetic induction is 2B₁ in the direction of the second field.")
When the currents flowing in the parallel wires are in opposite directions, we obtain two magnetic fields in the same direction at any midpoint as shown in the figure. Thus, for the first magnetic induction B₁ at the distance d/2 we have
B₁ = μ₀ ∙ I/2π ∙ d/2
= μ₀ ∙ I/π ∙ d
and the second magnetic induction B₂ is
B₂ = - μ₀ ∙ 3I/2π ∙ d/2
= 3μ₀ ∙ I/π ∙ d
= 3B₁
(Here, we have two negative signs: one for the position of the second wire in respect to the midpoint and the other for the direction of magnetic field produced by the second wire which is opposite to that of the first wire because currents are antiparallel.)
Therefore, the net magnetic induction at midpoint between the wires is
B_net = B₁ + B₂
= B₁ + 3B₁
= 4B₁

Since the intensity of magnetic field decreased with the increase in distance from the current carrying wire, we also decrease the intensity of magnetic field lines when representing it visually to give the idea of a weaker magnetic field when moving away from the wire as shown in the figure.

b. Magnetic Field of a Current-Carrying Loop

If a current flows in a circular loop as shown in the figure below, the four curled fingers are placed in the direction of electric current while the thumb shows the direction of magnetic field.

The magnetic field at the centre of a circular loop of radius r is perpendicular to the plane of the loop as shown in the figure.

Its magnitude is given by the formula:

B = μ₀ ∙ I/2r

If the loop has many turns (N turns), we obtain a flat coil with a single radius. Thus, we obtain for the magnetic field at centre of coil:

B = N ∙ μ₀ ∙ I/2r

Example 4

A circular loop containing 200 turns has a magnetic field of 0.2mT at its centre when a 5 A current Is flowing through it. What is the radius of loop?

Clues:

N = 200 turns = 2 × 102 turns
B = 20 mT = 2 × 10^-2 T
I = 5 A
(μ0 = 4π × 10^-7 N/A2)
r = ?

Solution 4

From the formula of magnetic field at centre of a loop with N turns, we have

B = N ∙ μ₀ ∙ I/2r

Rearranging for the radius r, we obtain

r = N ∙ μ₀ ∙ I/2B
= (2 × 10² ) ∙ (4 ∙ 3.14 × 10^-7 N/A² ) ∙ (5A)/2 ∙ (2 × 10^-2 T)
= 3.14 × 10^-2 m
= 3.14 cm

If there are two or more current carrying wires of different shapes, the magnetic field at a given position is the net magnetic field of each individual wire at the given point. Let's clarify this point through an example.

Example 5

Two antiparallel current carrying wires having magnitudes of 4A and 5A respectively are 20 cm apart as shown in the figure. The 4A wire has a semi-circular shape of radius 40 cm around the point O. Physics Tutorials: This image provides visual information for the physics tutorial Magnetic Field Produced by Electric Currents

What is the direction of magnetic field at the point O?
What is the net magnetic induction at this point?

Solution 5

Let's use the two right-hand rules discussed earlier to find the direction of each magnetic field. Thus, for the loop we curl the four fingers of the right hand in the direction of the 4A current and as a result, the thumb showing the direction of magnetic field will be directed inwards the page, while for the 5 A current we grasp the wire with the right hand where the thumb shows the current. Hence, the four curled fingers will show the direction of magnetic field. In this way, we find that both magnetic fields are normal to the page in the inwards direction (we must use the symbol ⊗ to show the direction for both magnetic fields).
The magnitude of the 4A wire at centre of the loop is
B₁ = 1/2 ∙ μ₀ ∙ I/2r
(The coefficient 1/2 is because the loop's shape is semi-circular.)
Hence, substituting the values, we obtain for the first magnetic induction (giving that r = 40 cm = 0.4 m):
B₁ = (4 ∙ 3.14 × 10^-7 N/A² ) ∙ (4A)/4 ∙ (0.4 m)
= 3.14 × 10^-6 T
As for the magnetic field produced by the 5A (straight) wire, we obtain
B₂ = μ₀ ∙ I/2π ∙ r
= (4 ∙ 3.14 × 10^-7 N/A² ) ∙ (5A)/2 ∙ 3.14 ∙ (0.4 m)
= 2.5 × 10^-6 T
Therefore, the net magnetic field at the point O is
B_net = B₁ + B₂
= 3.14 × 10^-6 T + 2.5 × 10^-6 T
= 5.64 × 10^-6 T
= 5.64 μT

c. Magnetic Field Produced by a Solenoid

As explained earlier, a solenoid is a spring-like conducting wire in which a current I flows. The wire is very long because the magnitude of magnetic field depends on the number of turns, as discussed in the previous paragraph.

The solenoid turns are assumed as closely packed to each other so that there are no spaces between any two adjacent turns, where each turn represents an individual coil.

The magnetic field outside the solenoid is weak and non-uniform, so we can focus mainly on the magnetic field inside the solenoid. If the solenoid's length L is much longer that the diameter of turns, the magnetic field lines are linear and parallel to each other, similar to the magnetic field inside a bar magnet (the external magnetic field is similar to that of a bar magnet as well). The direction of this magnetic field is found by applying the right hand rule. Thus, for the direction of current shown in the figure above, we determine the direction of the corresponding magnetic field.

In this way, one end of the solenoid acts like a north magnetic pole while the other end as a south magnetic pole, as shown in the figure.

The magnitude of magnetic induction (field) inside a solenoid of length L containing N turns when a current I flows through it, is

B = μ₀ ∙ N ∙ I/L

If we write n instead of N/L, where n represents the number of turns per unit length, the above equation becomes

B = μ₀ ∙ n ∙ I

From the above equation we conclude that magnetic field inside a solenoid is directly proportional to the number of turns per unit length and the amount of current flowing through the solenoid.

Example 6

A 5A current flows through a 30-cm long solenoid containing 500 turns. Calculate the magnetic induction inside the solenoid. Take the figure in the theory as a reference.

Solution 6

In this problem, we have the following clues:

I = 5 A
L = 30 cm = 0.30 m
N = 500 = 5 × 10²
(μ0 = 4π × 10^-7 N/A2)
B = ?

Using the equation of magnetic induction inside a solenoid,

B = μ₀ ∙ N ∙ I/L

we obtain after substitutions:

B = (4 ∙ 3.14 × 10^-7 N/A) ∙ (5 × 10² ) ∙ (5A)/0.30 m
= 1.047 × 10^-2 T

As you see, this value of magnetic field is considerably higher than the magnetic fields produced by individual straight wires discussed in the previous examples. Therefore, solenoid is a great tool used to obtain a strong magnetic field in a narrow space. You can find many examples of solenoids in electric appliances such as in inductors, electromagnets, antennas, valves and many more devices in which a strong magnetic field is required.

You have reached the end of Physics lesson 16.2.3 Magnetic Field Produced by Electric Currents. Right Hand Rule. There are 4 lessons in this physics tutorial covering Magnetic Field Produced by Electric Currents, you can access all the lessons from this tutorial below.

More Magnetic Field Produced by Electric Currents Lessons and Learning Resources

Magnetism Learning Material
Tutorial ID	Physics Tutorial Title	Revision Notes	Revision Questions
16.2	Magnetic Field Produced by Electric Currents
Lesson ID	Physics Lesson Title	Lesson	Video Lesson
16.2.1	Two Experiments to Prove the Interaction between Electricity and Magnetism
16.2.2	The Definition of Magnetic Induction
16.2.3	Magnetic Field Produced by Electric Currents. Right Hand Rule
16.2.4	Magnetic Permeability. Relative Permeability. Diamagnetic, Paramagnetic and Ferromagnetic Materials

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Physics Lesson 16.2.3 - Magnetic Field Produced by Electric Currents. Right Hand Rule

Magnetic Field Produced by Electric Currents. Right Hand Rule

a. Magnetic Field Produced by a Long Straight Wire

Example 2

Solution 2

Example 3

Solution 3

b. Magnetic Field of a Current-Carrying Loop

Example 4

Solution 4

Example 5

Solution 5

c. Magnetic Field Produced by a Solenoid

Example 6

Solution 6

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