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In this Physics tutorial, you will learn:

- How do electricity and magnetism interact with each other?
- What is the quantity used to represent the magnetic field? What kind of quantity is it (vector or scalar)?
- What are the factors affecting the magnetic induction? How to measure it?
- How to use the right-hand rule to determine the direction of magnetic field?
- How to represent symbolically the direction of magnetic field or that of current in 2D?
- How to find the direction and magnitude of a magnetic field produced by a long current-carrying wire?
- How to find the direction and magnitude of a magnetic field produced by a loop/ by a coil composed by a number of loops?
- How to find the direction and magnitude of a magnetic field produced by a solenoid?
- What is permeability of free space (vacuum)?
- What is relative permeability? How does it affect the properties of magnetic materials?
- How do we classify the magnetic materials?

B*⃗* = *F**⃗**/**I ∙ L*

From the above formula, we can determine the unit of magnetic induction B known as tesla, T. Thus, 1[T] = 1[*N**/**A ∙ m*]

In SI units, it becomes 1[T] = 1[*kg ∙ m/s*^{2}*/**A ∙ m*] = 1[*kg**/**A ∙ s*^{2}]

Magnetic induction B is the quantity that represents the magnetic field. It is analogue to the electric field E or gravitational field g discussed in the previous chapters. Example: What is the value of magnetic induction produced by a 3A current flowing through a conducting wire which is 20 cm long? The magnetic force produced in the given point is equal to 30 N. Solution Clues: F = 30 N L = 20 cm = 0.20 m I = 3 A B = ? From the definition of magnetic induction, we have B*⃗* = *F**⃗**/**I ∙ L*

=*30 N**/**(3 A) ∙ (0.20 m)*

=*50 N**/**A ∙ m*

= 50 T

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. 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 =

=

= 50 T

B*⃗* ∝ *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 μ_{0} = 4π × 10^{-7} N/A^{2}

Thus, when considering all the above factors, the formula of magnetic induction (in scalar form) produced by a long current carrying wire is B = (μ_{0} ∙ I)/(2π ∙ r)

Example: a) What is the magnetic induction at a distance of 40 cm from a long conducting wire carrying a 0.3 A current? b) What is the direction of magnetic field if the current is flowing vertically down? Solution Clues: r = 40 cm = 0.40 m I = 0.3 A μ0 = 4π × 10-7 N/A2 a) B = ? b) Direction of B = ? a) From the equation of magnetic induction in a long current carrying conducting wire, we have: B = (μ_{0} ∙ I)/(2π ∙ r) = (4π × 10^{-7} N/A^{2} ∙ 0.3 A)/(2π ∙ 0.40 m) = 0.75 × 10^{-7} N/(A ∙ m) = 7.5 × 10^{-8} T

b) 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: 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: a) The currents have the same direction b) The currents have opposite direction Solution a) 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 B_{1} = (μ_{0} ∙ I)/(2π ∙ d/2) = (μ_{0} ∙ I)/(π ∙ d)

The magnetic induction B2 at the distance - d/2 from the wire 3I is B_{2} = (μ_{0} ∙ 3I)/(2π ∙ (-d/2) ) = -(3μ_{0} ∙ I)/(π ∙ d) = -3B_{1}

(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 = B_{1}+B_{2} = B_{1}-3B_{1} = -2B_{1}

(Or we can say "the net magnetic induction is 2B1 in the direction of the second field.") b) 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 B_{1} = (μ_{0} ∙ I)/(2π ∙ d/2) = (μ_{0} ∙ I)/(π ∙ d)

and the second magnetic induction B2 is B_{2} = -(μ_{0} ∙ 3I)/(2π ∙ (-├ d/2┤)) = (3μ_{0} ∙ I)/(π ∙ d) = 3B_{1}

(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 = B_{1}+B_{2} = B_{1}+3B_{1} = 4B_{1}

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 = (μ_{0} ∙ 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 ∙ (μ_{0} ∙ I)/2r

Example: 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 From the formula of magnetic field at centre of a loop with N turns, we have B = N ∙ (μ_{0} ∙ I)/2r

Rearranging for the radius r, we obtain r = (N ∙ μ_{0} ∙ I)/2B = ((2 × 10^{2} ) ∙ (4 ∙ 3.14 × 10^{-7} N/A^{2} ) ∙ (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: 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. a) What is the direction of magnetic field at the point O? b) What is the net magnetic induction at this point? Solution a) 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). b) The magnitude of the 4A wire at centre of the loop is B_{1} = 1/2 ∙ (μ_{0} ∙ 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): B_{1} = ((4 ∙ 3.14 × 10^{-7} N/A^{2} ) ∙ (4A))/(4 ∙ (0.4 m) ) = 3.14 × 10^{-6} T

As for the magnetic field produced by the 5A (straight) wire, we obtain B_{2} = (μ_{0} ∙ I)/(2π ∙ r) = ((4 ∙ 3.14 × 10^{-7} N/A^{2} ) ∙ (5A))/(2 ∙ 3.14 ∙ (0.4 m) ) = 2.5 × 10^{-6} T

Therefore, the net magnetic field at the point O is B_net = B_{1}+B_{2} = 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 = (μ_{0} ∙ 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 = μ_{0} ∙ 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: 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 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 = (μ_{0} ∙ N ∙ I)/L

we obtain after substitutions: B = ((4 ∙ 3.14 × 10^{-7} N/A) ∙ (5 × 10^{2} ) ∙ (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. Magnetic Permeability. Relative Permeability. Diamagnetic, Paramagnetic and Ferromagnetic Materials In all formulae discussed in the previous paragraphs, the effect of the surrounding medium is assumed as negligible. This occurs when this medium is the vacuum. We included in all formulae the effect of vacuum by inserting the quantity known as vacuum magnetic permeability (or magnetic permeability of free space), which value is μ0 = 4π × 10-7 N/A2. What if the magnetic field is not produced in vacuum but in another medium instead? In this case, we have to consider the effect of the other medium as well. We don't expect the magnetic field have the same value as if there was vacuum, due to the presence of extra matter, which affects the magnitude of magnetic field. Such a presence is represented through a quantity known as relative permeability, μr, which is calculated by μ_r = B/B_{0} = μ/μ_{0}

where B is the magnetic field in presence of a certain matter, B0 is the corresponding magnetic field in vacuum; μ and μ0 are the values of permeability in the presence of the given matter and in vacuum respectively. The values of relative magnetic permeability of some materials are given in the table below. If we look carefully the table above, we observe in the list three kind of materials: 1- The first category includes materials that have a relative permeability slightly smaller than 1. They are known as diamagnetic materials. The magnetic field produced in them is slightly weaker than the magnetic field produced in the same circumstances but in vacuum. Examples of diamagnetic materials include carbon, bismuth, silver, copper, etc. When a diamagnetic is placed inside a magnetic field, it is weakly magnetised in the opposite direction of magnetic field. As a result, the overall magnetic field decreases. Therefore, diamagnetic materials are used for decreasing the magnetic field of a given magnet when necessary. This means the magnetic field lines diverge from each other when approaching the material as shown in the figure below. 2- The second category includes materials that have a relative permeability slightly higher than 1. They are known as paramagnetic materials. Examples of paramagnetic materials include aluminium, magnesium, air, etc. When a diamagnetic is placed inside a magnetic field, it is slightly magnetised in the direction of this field. Therefore, the magnetic field lines slightly converge when approaching the material. However, this convergence cannot be seen at naked eye as it is too small to be noticeable and as a result, the magnetic field lines are parallel to each other as shown in the figure below. 3- Ferromagnetic materials are those materials that have their relative permeability much greater than 1 (a few hundred to a few thousand times greater than 1). They are strongly magnetised when placed inside a magnetic field. As a result, the magnetic field enforces and the field lines converge when approaching the material, as shown in the figure below. Iron cobalt and nickel are examples of ferromagnetic materials. Due to their strong magnetic properties, ferromagnetic materials are used to produce artificial magnets. Summary An electric current produces a magnetic field in the space around it and magnetism generates electricity when a magnet is moving towards to or away from a coil in which a current is flowing. These two properties form the foundations of electromagnetism. Magnetic induction is the quantity that represents the magnetic field. It is analogue to the electric field E or gravitational field g. Magnetic induction is a vector quantity and is denoted by BB*⃗* = F*⃗*/(I ∙ L)

The unit of magnetic induction B known as tesla, T. 1[T] = 1[N/(A ∙ m)]. In SI units, 1[T] = 1[kg/(A ∙ sB = (μ_{0} ∙ I)/(2π ∙ r)

where μB = (μ_{0} ∙ I)/2r

where r is the radius of the loop. 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 ∙ (μ_{0} ∙ I)/2r

The magnitude of magnetic induction (field) inside a solenoid (a spring-like conducting wire) of length L containing N turns when a current I flows through it, is B = (μ_{0} ∙ 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 = μ_{0} ∙ n ∙ I

The above equation is a demonstration 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. The presence of extra matter, which affects the magnitude of magnetic field is represented through a quantity known as relative permeability, μr, which is calculated by μ_r = B/B_{0} = μ/μ_{0}

where B is the magnetic field in presence of a certain matter, B0 is the corresponding magnetic field in vacuum; μ and μ0 are the values of permeability in the presence of the given matter and in vacuum respectively. There are three types of magnetic materials: 1- Diamagnetic materials. They have a relative permeability slightly smaller than 1. When a diamagnetic is placed inside a magnetic field, it is weakly magnetised in the opposite direction of magnetic field. As a result, the overall magnetic field decreases and the magnetic field lines diverge from each other when approaching the material. Carbon, bismuth, silver and copper are examples of diamagnetic materials. 2- Paramagnetic materials. This category includes materials that have a relative permeability slightly higher than 1. Examples of paramagnetic materials include aluminium, magnesium, air, etc. When a diamagnetic is placed inside a magnetic field, it is slightly magnetised in the direction of this field. Therefore, the magnetic field lines slightly converge when approaching the material. 3- Ferromagnetic materials are those materials that have their relative permeability much greater than 1 (a few hundred to a few thousand times greater than 1). They are strongly magnetised when placed inside a magnetic field. As a result, the magnetic field enforces and the field lines converge when approaching the material. Iron cobalt and nickel are examples of ferromagnetic materials. Due to their strong magnetic properties, ferromagnetic materials are used to produce artificial magnets. **1)** Two parallel wires carrying currents of I1 = 5A and I2 = 10A flowing in the same direction are placed 10 cm apart as shown in the figure.

What is the magnitude and direction of magnetic field at the point A which is 5 cm on the left of the 5A current?

- 3.33 × 10-5 T clockwise
- 10-4 T anticlockwise
- 3.33 × 10-5 T anticlockwise
- 1.047 × 10-4 clockwise

**Correct Answer: C**

**2)** A circular loop of radius 4 cm containing 100 turns has a magnetic field of 0.2mT at its centre when some current is flowing through it. What is the value of current in amps?

- 127 A
- 12.7 A
- 1.27 A
- 0.127 A

**Correct Answer: D**

**3)** A 2A wire is bent in the shape shown in the figure.

What is the magnetic field at centre of the loop if its radius is 12 cm?

- 3.33 × 10-6 T
- 1.38 × 10-5 T
- 3.76 × 10-5 T
- 7.13 × 10-6 T

**Correct Answer: B**

We hope you found this Physics tutorial "Magnetic Field Produced by Electric Currents" 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 our Physics tutorial on Magnetic Force on a Current Carrying Wire. Ampere's Force.

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