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In this Physics tutorial, you will learn:

- What is Ampere's Force? How it is calculated?
- How to determine the direction of magnetic force produced when a current carrying wire is inserted inside a magnetic field?
- How to calculate the magnetic force produced between two current carrying wires?
- What happens when a current carrying loop is inserted inside a uniform magnetic field?
- What is electric motor and how does it work?

What effect does a magnetic field cause on magnetic materials? Why does this occur?

How does the interacting ability of magnetic materials changes when we increase the strength of magnetic field? What about current? How does it affect this ability?

In this tutorial, we will discuss about the magnetic force caused by a current carrying wire. It is a known fact that force is a vector quantity, i.e. it has a direction. This means besides the calculation of magnetic force, you will learn how to determine its direction as well. This is particularly useful when trying to operate dynamic devices using electromagnetism such as electric motors.

From the property of interaction between electricity and magnetism discussed in the previous tutorial, it is clear that a conducting wire will not produce any magnetic field when no current flows through it. On the other hand, a static magnetic field will not produce any current If no conducting wire is placed inside the field. Therefore, a simultaneous existence of a magnetic field and electric current Is a necessity to cause an interaction between them.

The term "interaction" is nothing else but another word to express the existence of a force. We know from the previous tutorial (from the definition of magnetic induction B) that the magnetic force F produced on a conducting wire of length L when a current I flows through it, is:

F*⃗* = I ∙ ( L*⃗* × B*⃗* )

This is a mixed product of vectors, which has the following corresponding scalar form:

F = I ∙ L ∙ B ∙ sinθ

where θ is the angle between the current carrying wire and the direction of magnetic field and I, L and B are the magnitudes of the corresponding quantities regardless their direction. This equation (either in scalar or vector form) is known as the equation of **Ampere's Force**.

The direction of magnetic force calculated through the formula of Ampere's Force is found by using the so-called "**Fleming's Left Hand Rule**". It consists of three steps:

- The left hand is completely open; the thumb forms an angle of 90° to the other four fingers.
- The four fingers show the current direction.
- The magnetic field lines punch the open palm.

If these three rules are applied properly, then the thumb shows the direction of magnetic force, as shown in the figure.

There are two other versions of left hand rule application as well. One is the three fingers version (FBI) where the thumb shows the force, the index finger the direction of magnetic field and the middle finger the current, as shown in the figure below.

The other version is the following one:

Henceforth we will use the first version of Fleming's LHR in all situations to avoid confusion.

The electric circuit shown in the figure below produces a 3A current when the switch is closed. The circuit is placed normal to a uniform magnetic field of induction equal to 5 mT. A 20 cm metal bar is placed perpendicular with the two horizontal circuit sides. The switch is initially open. Then, it turns ON for a very short time and is switched off again.

- What is the moving direction of the metal bar immediately after the switch is open?
- What is the magnitude of magnetic force produced on the bar? Suppose almost all the current flows only through the bar when the circuit is open.
- What is the mass of the bar if it starts accelerating from rest at 0.2 m/s2 when the switch turns on?

- The cross symbols mean the magnetic field lines enter the page normal to it. Thus, the palm must be directed out of page.

Electric current flows in the anticlockwise direction when the switch is open. This means the current in the bar flows in the down-up direction, so the four fingers will be directed upwards.

From here, we can find the direction of the magnetic force applying the Fleming's Left Hand Rule. Putting the hand in the required direction, we find that the magnetic force shown by the thumb acts due left.

- Clues:

I = 3A

B = 5 mT = 0.005 T

L = 20 cm = 0.20 m

F = ?

From the formula derived from the Fleming's Left Hand Rule (the scalar version), we obtainF = I ∙ L ∙ B ∙ sinθwhere sin θ = 1 because all components of the formula are perpendicular to each other (sin 90° = 1). Thus, substituting the values, we obtainF = (3A) ∙ (0.20m) ∙ (0.005T) ∙ 1

= 0.003 N - From the Newton's Second Law of Motion, we have a =where m is the mass of the conducting rod, F is the magnetic force acting on the rod and a is its acceleration. Therefore, rearranging the above formula we obtain for the mass m of the rod:
*F**/**m*m =*F**/**a*

=*0.003 N**/**0.2 m/s*^{2}

= 0.015 kg

= 15 g

The same procedure is also used when the angle between the current and magnetic field lines is not 9°. Let's consider an example.

A current carrying bar is used to connect two detached parts of an electric circuit as shown in the figure.

What are the horizontal and vertical components of magnetic force acting on the bar if the value of magnetic induction is 20 mT?

The direction of magnetic field is out of the page, normal to its plane. In addition, the value of electric current flowing through the circuit (and therefore through the metal bar) is calculated applying the Ohm's Law where ε = 12 V and R = 30 Ω. Therefore, we obtain for current:

I = *ε**/**R*

=*12 V**/**30 Ω*

= 0.4 A

=

= 0.4 A

The direction of magnetic field is still perpendicular to the current, regardless the angle shown in the figure. This is because the electric current lies according the plane of the figure while magnetic field is normal to it. Therefore, we have for the magnitude of magnetic force produced:

F = I ∙ L ∙ B ∙ sinθ

where sin θ = 1 because L and B of are perpendicular to each other (sin 90° = 1). Thus, giving that L = 10 cm = 0.10 m and B = 20 mT = 0.020 T, we obtain after substitutions

F = (0.4 A) ∙ (0.10 m) ∙ (0.020 T) ∙ 1

= 0.0008 N

= 0.0008 N

Using the Fleming's left hand rule (the palm is directed onto the page, the four fingers are in the up-right direction) the magnetic force shown by the thumb is directed inwards the circuit perpendicular to the bar. The horizontal component F_{x} of the magnetic force is

F_{x} = F ∙ cos(90^{0} - 53^{0} )

= 0.0008 N ∙ cos37^{0}

= 0.0008 N ∙ 0.8

= 0.00064 N (due right)

= 0.0008 N ∙ cos37

= 0.0008 N ∙ 0.8

= 0.00064 N (due right)

and the vertical component F_{y} of the magnetic force is

F_{y} = F ∙ sin(90^{0} - 53^{0} )

= 0.0008 N ∙ sin37^{0}

= 0.0008 N ∙ 0.6

= 0.00048 N (down)

= 0.0008 N ∙ sin37

= 0.0008 N ∙ 0.6

= 0.00048 N (down)

The figure below shows these magnetic force components.

Since the interaction between magnetic field and electric current brings a magnetic force on the wire, we will have a mutual interaction if two current carrying wires are placed inside a magnetic field. The simplest case in this regard involves two parallel current carrying conductors inside a magnetic field where currents lie in the same direction as shown in the figure.

The two parallel wires having lengths L_{1} and L_{2} respectively are separated by a distance d from each other. In addition, the magnetic fields they produce are B_{1} and B_{2} respectively. As a result, a magnetic force F_{12} produced by the first wire will act on the second wire while the magnetic force F_{21} produced by the second wire acts on the first one. These forces are equal and opposite based on the action-reaction principle (Newton's Third Law of Motion). You can also apply the Fleming's Left Hand Rule to convince yourself that these forces have opposite directions.

As for the magnetic fields, we can use the following reasoning:

The first wire (which carries a current of I_{1}) produces a magnetic field B_{12} at the position of the second wire (at distance d from the first wire). The magnitude of B_{12} produced by the first wire at the position of the second wire therefore is

B_{12} = *μ*_{0} ∙ I_{1}*/**2π ∙ d*

and the corresponding magnetic force F_{12} of the first wire on the second, is

F_{1}2 = I_{2} ∙ B_{12} ∙ L_{2}

=*μ*_{0} ∙ I_{1} ∙ I_{2}*/**(2π ∙ d) ∙ L*_{2}

=

Likewise, the magnetic force F_{21} by which the second wire acts on the first, is

F_{2}1 = I_{1} ∙ B_{21} ∙ L_{1}

=*μ*_{0} ∙ I_{1} ∙ I_{2}*/**(2π ∙ d) ∙ L*_{1}

=

If the two wires have the same length and current, the magnitudes of the two above forces are equal. Since forces have opposite directions, the wires repel each other when parallel currents flow in them. On the other hand, when currents are antiparallel, i.e. when they have opposite directions, the two wires attract each other based on the direction of magnetic forces produced. Look at the figure:

Two identical parallel bars are placed horizontally one above the other as shown in the figure.

The bars are 50 cm long each. The lower bar carries a current of 800A and it is unmoveable, while the upper bar (the 500A bar) can move freely when released and it is 20 cm above the lower bar. What is the mass of the upper bar so that it will stay in the actual position after being released? Take g = 9.81 m/s2.

Clues:

I_{1} = 800A = 8 × 10^{2} A

I_{2} = 500A = 5 × 10^{2} A

L_{1} = L_{2} = 50 cm = 0.50 m

d = 20 cm = 0.20 m

(g = 9.81 m/s^{2})

(μ_{0} = 4π × 10^{-7} N/A^{2})

m_{2} = ?

The condition for the upper bar to stay unmoveable in the actual position after release is that magnetic force by which the lower bar acts on it, be equal to the weight of the upper bar, i.e.

F_{12} = W_{2}

*μ*_{0} ∙ I_{1} ∙ I_{2}*/**2π ∙ d* ∙ L_{2} = m_{2} ∙ g

m_{2} = *μ*_{0} ∙ I_{1} ∙ I_{2} ∙ L_{2}*/**2π ∙ d ∙ g*

=*(4 ∙ 3.14 × 10*^{-7} *N**/**A*^{2} ) ∙ (8 × 10^{2} A) ∙ (5 × 10^{2} A) ∙ (0.50m)*/**2 ∙ 3.14 ∙ (0.20m) ∙ (9.81 **m**/**s*^{2})

= 20.4 × 10^{-3} kg

= 20.4 g

m

=

= 20.4 × 10

= 20.4 g

Let's consider a rectangular loop placed between the opposite poles of two magnets as shown in the figure.

A current I flows through the loop as shown by the yellow arrows. The magnetic field lines lie from North to South poles of magnets (from right to left). Using the Fleming's Left Hand Rule (magnetic field lines punch the palm, the four fingers show the current, the thumb shows the magnetic force), we see that a torque is produced in the two lateral sides of the loop because current flows in opposite directions.

To calculate the torque on the loop we must calculate the forces acting at the lateral sides of the loop only because the front and the back sides are parallel to the magnetic field lines and therefore, they don't produce any force (sin 00 = 0, so F = I ∙ B ∙ L ∙ sin 00 = 0). Let's denote the four vertices of the loop by a, b, c and d respectively. The turning arm is therefore 'ad'/2 or 'bc'/2. If we denote by L the loop sides ab and bc in which the magnetic force produces rotation, we obtain for the scalar version of magnetic forces F_{1} and F_{2}:

F_{1} = I ∙ B ∙ ('ab') ∙ sin90^{0}

= I ∙ B ∙ L

= I ∙ B ∙ L

and

F_{2} = I ∙ B ∙ ('cd') ∙ sin90^{0}

= I ∙ B ∙ L

= I ∙ B ∙ L

Therefore, the magnetic forces F_{1} and F_{2} are equal in magnitude and opposite in direction. If we denote by ** x/2** the distance from the lateral sides ab and bc to the rotating axis passing through the centre of loop, we obtain for the maximum torque τ produced:

τ_{max} = F_{1} ∙ *x**/**2* + F_{2} ∙ *x**/**2*

= I ∙ B ∙ L ∙*x**/**2* + I ∙ B ∙ L ∙ *x**/**2*

= 2I ∙ B ∙ L ∙*x**/**2*

= I ∙ B ∙ L ∙ x

= I ∙ B ∙ L ∙

= 2I ∙ B ∙ L ∙

= I ∙ B ∙ L ∙ x

where x is the width of the loop (it represents the side lengths ad or bc).

Since L ∙ x gives the area A of the loop, we can write the last equation as

τ_{max} = I ∙ B ∙ A

What is the maximum torque produced by the rectangular loop placed between the opposite poles of two magnets shown in the figure? The magnetic field between the opposite poles is 5mT and the current flowing through the loop is 4A. The dimensions of the loop are 30cm × 20cm. What is the turning direction of the loop?

Clues:

I = 4A

B = 5mT = 5 × 10-3 T

A = 30 cm × 20 cm = 600 cm2 = 0.06 m2 = 6 × 10^{-2} m^{2}

τ_{max} = ?

The maximum torque produced by this system is

τ_{max} = I ∙ B ∙ A

= (4A) ∙ (5 × 10^{-3} T) ∙ (6 × 10^{-2} m^{2} )

= 120 × 10^{-5} N ∙ m

= 0.0012 N ∙ m

= (4A) ∙ (5 × 10

= 120 × 10

= 0.0012 N ∙ m

Electric motor uses the interaction between electric current and magnetic field to produce motion. The example with the current carrying coil placed between two magnets discussed earlier represents the main mechanism of an electric motor. Basically, an electric circuit supplies the coil with electricity, while the magnetic field is provided by the two magnets between which the coil is placed. As a result, a movement in the form of torque (as seen in the examples of the previous paragraph) is produced.

There is a drawback to overcome in this mechanism however. The electric circuit which supplies the coil, cannot rotate together with the coil as it rotates very fast. The circuit must remain stationary while the coil rotates. This is made possible by using a commutator (the cylindrical part shown in the figure in theory)

The output of electric motor is mechanical energy. Thus, we can say: "**an electric motor is a device that converts electricity into mechanical energy.**"

Many electric appliances such as washing machines, vacuum cleaners, fans, air conditioners, refrigerators, etc., make use of electric motors to operate. In addition, there exist many other industrial appliances which use electric motors among which there are pumps, compressors, crushers, HVAC systems, etc.

The magnetic force F produced on a conducting wire of length L when a current I flows through it, is:

F*⃗* = I ∙ ( L*⃗* × B*⃗* )

where B is the magnetic field produced during this process.

This is a mixed product of vectors, which has the following corresponding scalar form:

F = I ∙ L ∙ B ∙ sinθ

where θ is the angle between the current carrying wire and the direction of magnetic field and I, L and B are the magnitudes of the corresponding quantities regardless their direction. This force calculated either in scalar or vector form is known as the **Ampere's Force**.

The direction of magnetic force calculated through the formula of Ampere's Force is found by using the so-called "**Fleming's Left Hand Rule**". It consists of three steps:

- The left hand is completely open; the thumb forms an angle of 90° to the other four fingers.
- The four fingers show the current direction.
- The magnetic field lines punch the open palm.

If these three rules are applied properly, then the thumb shows the direction of magnetic force.

If we have two parallel wires of length L carrying currents I, then the magnitude field B_{12} produced by the first wire at the position of the second wire is

B_{12} = *μ*_{0} ∙ I_{1}*/**2π ∙ d*

and the corresponding magnetic force F_{12} of the first wire on the second, is

F_{12} = I_{2} ∙ B_{1}2 ∙ L_{2}

=*μ*_{0} ∙ I_{1} ∙ I_{2}*/**(2π ∙ d) ∙ L*_{2}

=

Likewise, the magnetic force F_{21} by which the second wire acts on the first, is

F_{21} = I_{1} ∙ B_{2}1 ∙ L_{1}

=*μ*_{0} ∙ I_{1} ∙ I_{2}*/**(2π ∙ d) ∙ L*_{1}

=

If the two wires have the same length and current, the magnitudes of the two above forces are equal. Since forces have opposite directions, the wires repel each other when parallel currents flow in them. On the other hand, when currents are antiparallel, i.e. when they have opposite directions, the two wires attract each other based on the direction of magnetic forces produced.

A current-carrying loop placed inside the magnetic field produced by two magnets with opposite poles facing each other, produces a torque when inserted properly. This is because the current flows in opposite directions in the two lateral sides of the loop. Therefore, the magnetic forces F_{1} and F_{2} are equal in magnitude and opposite in direction. If we denote by ** x/2** the distance from the lateral sides ab and bc to the rotating axis passing through the centre of loop, we obtain for the maximum torque τ produced:

τ_{max} = I ∙ B ∙ L ∙ x = I ∙ B ∙ A

where A = L ∙ x is the area of loop.

Electric motor uses the interaction between electric current and magnetic field to produce motion. Basically, an electric circuit supplies the coil with electricity, while the magnetic field is provided by the two magnets between which the coil is placed. As a result, a movement in the form of torque is produced.

The output of electric motor is mechanical energy. Thus, we can say: "**an electric motor is a device that converts electricity into mechanical energy**."

Many electric appliances such as washing machines, vacuum cleaners, fans, air conditioners, refrigerators, etc., make use of electric motors to operate. In addition, there exist many other industrial appliances which use electric motors among which there are pumps, compressors, crushers, HVAC systems, etc.

**1)** A 12 cm long metal bar is suspended by two identical springs of constants equal to 4 N/m each. This system is inserted inside a 50 mT magnetic field directed onto the page. If the mass of the bar is negligible and a 20 A current Is flowing through it, calculate the maximum extension produced in the springs. Remember the equivalent spring constant k of a parallel system of springs is k = k1 + k2.

- 1.5 cm
- 3 cm
- 6 cm
- 12 cm

**Correct Answer: A**

**2)** Two identical current carrying metal bars being 8 cm long each are used to connect the detached parts of an electric circuit supplied by a 12 V battery. The wires are parallel, placed 10 cm apart as shown in the figure.

What magnetic force is acting between the bars if the total resistance of the circuit is 20 ohms?

- 9.33 × 10-8 pushing force
- 5.76 × 10-8 pushing force
- 5.76 × 10-8 pulling force
- 9.33 × 10-8 pulling force

**Correct Answer: C**

**3)** A 8A current carrying rectangular coil is placed inside two magnets the opposite poles of which face each other. The coil fits perfectly the dimensions of the space between the two magnets as shown in the figure.

What is the direction of magnetic field between the magnets if the maximum torque produced is 3 × 10-3 N ∙ m?

- 25 T
- 2.5 T
- 2.5 mT
- 25 mT

**Correct Answer: D**

We hope you found this Physics tutorial "Magnetic Force on a Current Carrying Wire. Ampere's Force" useful. If you thought the guide useful, it would be great if you could spare the time to rate this tutorial and/or share on social media, this helps us identify popular tutorials and calculators and expand our free learning resources to support our users around the world have free access to expand their knowledge of physics and other disciplines. In our next tutorial, we expand your insight and knowledge of Magnetism with ourPhysics tutorial on Magnetic Force on a Wire Moving inside a Magnetic Field. Lorentz Force.

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