Theory of Solenoids, Relays, Magnetomotive Force and Magnetic Field Strength

Magnetomotive Force and Magnetic Field Strength:
 
If a length of conductor is wound to make multiple current-carrying loops side-by-side, the magnetic fields related with the individual loops join to form magnetic field acting lengthways via the coil as shown in figure below. This magnetic field is present whereas current flows via the conductor however collapse when the current ceases to flow when the source of electricity is eliminated.

2095_18.1.jpg

Figure: A Wound Coil generating an internal magnetic field

When a piece of magnetisable material is positioned in this field, it will experience a force termed to as a Magnetomotive Force. The force experienced will of course base on the intensity or strength of magnetic field. The magnetomotive force must not be thought of as a physical Newtonian force, (although it frequently results in this), but instead as analogous to the electromotive force of a battery or electrical supply. In this situation, the magnetic flux is thought of as analogous to the current flowing in the loop in an electric circuit.

Magnetomotive Force is formally stated as the force tending to persuade alignment of the magnetic domains in a material to form lines of flux beneath the influence of a magnetic field.

1472_18.2.jpg

Figure: A Closely Wound Coil and its related magnetic field

The coil shown in figure above comprises of N closely wound turns with the conductor fed with a current I, and hence the magnetic field created in the coil is continuous and uniform. A piece of magnetisable, or previously magnetised, material is positioned within the region within the coil. The magnetomotive force experienced by the material is directly proportional to current, I, producing it and the number of turns, N, of the coil.

Magnetomotive Force is given the symbol of MMF and has SI Units of Amps (A)

Magnetomotive force MMF = current x no. of Turns = NI (A)

Strictly, the unit of current is Amperes, in the SI system. This, though, tends to hide the fact that the force depends on the number of turns of the coil and so the units are frequently stated as Ampere-turns (At), where the number of turns is in the consequence dimensionless.

Magnetic Field Strength is stated as the magnetomotive force experienced per unit length in the uniform magnetic field.

Magnetic Field Strength consists of the symbol H and has units of Amps/metre (Am-1)

Magnetic Field Strength = Magnetomotive force/Length of coiled path, H = NI/d Am-1

Where d is the distance over which, the field in the coil is generated, that is, length of the wound coil.

Permeability:

There is as well an intrinsic relationship between Magnetic field strength and Magnetic flux Density. This is characterised by the property termed as Permeability.

Permeability is defined as the degree of magnetisation that a material undergoes in response to an applied magnetic field. This is the ratio of Magnetic Flux Density to the Magnetic Field Strength. This is analogous to resistance in the electrical circuit.

Permeability is represented by the symbol µ and has units of Henrys/metre (Hm-1)

Permeability = Magnetic Flux density/Magnetic field strength

Μ = B/H Hm-1

Permeability of Free Space: This is defined as the ratio of Magnetic Flux Density to Magnetic Field Strength evaluated at a distance of 1m from a long straight conductor carrying a current of 1A as shown in figure below. It is designated as µ0 = 4π x 10-7 = 12.57 x 10-7 Hm-1.

2416_18.3.jpg

Figure: Permeability of Free Space

The permeability of various materials is generally specified as relative permeability, µr , that is, relative to free space and hence:

Permeability μ = µr μo Hm-1

From this, the Magnetic Flux density of the field in the coil can be found as:

Magnetic Flux density B = μH = μ (NI/d) Wbm-2

If the wound coil consists of air as the core then μ = μ0. If, on the other hand, the core is a magnetic material like Ferrite, then the relative permeability of this material should be employed.

Solenoids:

The principle of making a magnetic field in a coil is exploited in the Solenoid. A solenoid is formed by winding a coil on an insulated former. A magnetic or magnetisable shaft is then inserted into the former and hence it is subjected to the magnetic field that is formed when the coil is energised, that is, when a current flows via it. Generally, one end of the shaft protrudes from the former. The magnetised shaft has South and North poles that can be aligned in either direction with respect to the field related with the coil as shown in figure below.

Whenever the coil is energised, the shaft will move beneath the influence of magnetic field in a direction determined by the rule for poles and hence the North Pole of the shaft moves in the direction of South Pole of the field related with the energised coil. Alternatively, the South Pole of the shaft might be attracted in the direction of the North Pole of the field related with the coil, depending on the orientation of both.

64_Motion of the shaft in the Magnetic field of a Solenoid.jpg
Figure: Motion of the shaft in the Magnetic field of a Solenoid

The end of the shaft protruding from the previous can be joined to a mechanical actuator, that is, a mechanical mechanism that causes certain kind of physical action or movement. This will be forced to move in similar direction as the shaft whenever the solenoid is energised. Different mechanical configurations can be constructed to accomplish a diversity of tasks. Frequently the solenoid is spring-loaded as shown in figure above and hence the shaft returns to a neutral position whenever the coil is de-energised, that is, when the solenoid is deactivated. The solenoids come in a variety of sizes, shapes and power ratings. They are generally used in door-lock mechanisms, pressure and flow control pumps, valves and many other industrial applications.

Linear Action Solenoids:

In its simplest form a shaft can be inserted into solenoid and the protruding end joined to a mechanism that needs a linear push or pull action.

Lever Action Solenoids:

Several mechanical devices need a lever action or even a rotary twisting action to produce a bending force. Different mechanisms and forms of levers can be designed as shown in figure below. All of them, though, ultimately rely on being driven by the end of mechanism inserted into longitudinal field of the coil.

Valves and Actuators:

The solenoid coil can be built in as portion of the mechanical device which it is employed to drive, like in the valves and actuators shown in figure below.

The solenoid principle can be extended to build a relay. This is a device which, when current flows via the coil of the solenoid, causes a mechanical arm to move and close or open electrical contacts. Such contacts are employed to feed power to a load that can be electrically isolated from the circuit driving relay. Therefore, using a relay driven by a low-power, solid-state switch such as a transistor operating from a low-voltage supply, high voltage or current and heavy power like mains electricity can be switched on an off by other electronic circuits or equipment as shown in figure below.

275_18.9.jpg

Figure: A Relay Driven by a Transistor Switch

A relay might contain more than one set of contacts, all activated by similar solenoid and switching circuit as shown. The contacts can be of simple ON/OFF type, which either close or open when the relay is activated, or they might be of the CHANGEOVER kind where one contact is closed and one is open whenever the relay is inactive and this condition is reversed whenever the relay is activated. The selection of relays is as shown in figure below.

Example: The coil of solenoid employed in an automatic door-lock is wound on a former of length 15cm and diameter 2.5cm. This is fed with a current of 100mA to produce an internal magnetic field strength of 250Am-1. Find out the number of turns needed in the coil and the total length of wire employed.

Solution:

Magnetic Field strength = H = NI/d = 250 Am-1

Then, N = Hd/I = (250 x 0.15)/0.1 = (2.5 x 15)/0.1 = 37.5/0.1

And hence, N = 375 turns

Length of the wire can be found as the number of turns times the diameter π. Then:

Length of wire = N Π 2 r = 375 x 3.14 x 0.025

Length of wire = 29.43 m

Latest technology based Electrical Engineering Online Tutoring Assistance

Tutors, at the www.tutorsglobe.com, take pledge to provide full satisfaction and assurance in Electrical Engineering help via online tutoring. Students are getting 100% satisfaction by online tutors across the globe. Here you can get homework help for Electrical Engineering, project ideas and tutorials. We provide email based Electrical Engineering help. You can join us to ask queries 24x7 with live, experienced and qualified online tutors specialized in Electrical Engineering. Through Online Tutoring, you would be able to complete your homework or assignments at your home. Tutors at the TutorsGlobe are committed to provide the best quality online tutoring assistance for Electrical Engineering Homework help and assignment help services. They use their experience, as they have solved thousands of the Electrical Engineering assignments, which may help you to solve your complex issues of Electrical Engineering. TutorsGlobe assure for the best quality compliance to your homework. Compromise with quality is not in our dictionary. If we feel that we are not able to provide the homework help as per the deadline or given instruction by the student, we refund the money of the student without any delay.