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|16.2||Magnetic Field Produced by Electric Currents|
In this Physics tutorial, you will learn:
In the previous tutorial, we discussed some general features of magnetism among which the concept of magnetic field and magnetic field lines. Now, we will extend the discussion about magnetic field including a quantitative approach, that is to make the user able calculate the magnetic field in various situations.
In 1819, Oersted - a Danish teacher - when performing some experiments in his laboratory, observed a deflection of compass needle to the N-S direction in presence of electric current as shown in the figure below.
Thus, in absence of electric current In the circuit (figure 1), he noticed that the compass needle was directed towards the earth magnetic poles. When he turned the switch on, Oersted noticed a deflection in the compass needle (figure 2).
This experiment is a demonstration that electricity causes a magnetic effect, i.e. an electric current produces a magnetic field around it.
The reverse is also true. A few years later, in 1830, Michael Faraday noticed that an electric current Is produced on a closed conducting loop when a magnet is moved towards or away the loop as shown in the figure below. Furthermore, the current produced is not produced due to direct contact but rather, due to induction (induced current) and it opposes the cause, i.e. the direction of the induced current Is opposite to the magnet direction of motion.
The spring-like conducting wire in the figure is known as a solenoid. It is made up by a number of turns known as coils. A single coil may have a circular or rectangular shape and it is different from a simple circular or rectangular metal frame as the term coil is used when some current flows through the metal frame.
This experiment is a proof that magnetism generates electricity. Therefore, magnetism and electricity are closely related to each other. If we study them separately, this is only for simplicity.
From the two experiments above (especially the second one), it is obvious that magnetic field is produced by moving electric charges. Since the current produced in this case is an induced current, we use a quantity called magnetic induction to represent quantitatively the magnetic field. Magnetic induction is a vector quantity and is denoted by B⃗ in formulae. It depends on three factors:
Mathematically, we have:
From the above formula, we can determine the unit of magnetic induction B known as tesla, T. Thus,
In SI units, it becomes
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.
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.
F = 30 N
L = 20 cm = 0.20 m
I = 3 A
B = ?
From the definition of magnetic induction, we have
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.
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
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
Thus, when considering all the above factors, the formula of magnetic induction (in scalar form) produced by a long current carrying wire is
r = 40 cm = 0.40 m
I = 0.3 A
μ0 = 4π × 10-7 N/A2
a) B = ?
b) Direction of B = ?
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:
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.
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:
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:
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?
N = 200 turns = 2 × 102 turns
B = 20 mT = 2 × 10-2 T
I = 5 A
(μ0 = 4π × 10-7 N/A2)
r = ?
From the formula of magnetic field at centre of a loop with N turns, we have
Rearranging for the radius r, we obtain
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.
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.
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
If we write n instead of N/L, where n represents the number of turns per unit length, the above equation becomes
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.
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.
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,
we obtain after substitutions:
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.
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
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:
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