ELECTROMAGNETISM
When a conductor carries an electric current, a magnetic field is produced around that conductor. When a wire/conductor carrying an electric current is placed above a magnetic needle, the needle will deflect clockwise or anticlockwise depending upon the direction of the current. If the current is in the same direction as the magnetic field, the needle will deflect clockwise. On the other hand, if the needle is in opposite direction to the magnetic field, the needle will deflect anticlockwise.
Relationship between the direction of current flow and the magnetic field
1) Corkscrew/Screw Rule
In order that the screw may travel in the same direction as the current, it has to be turned in the direction of the current (clockwise when viewed from the left – hand side)
2) Grip Rule:
The conductor is gripped with the right hand, with the thumb parallel to the conductor and pointing in the direction of the current. The fingers then points in the direction of the magnetic flux around the conductor.
Magnetic field of a solenoid
If a coil is wound on a steel rod and connected to a battery, the steel becomes magnetized and behaves like a permanent magnet. When a current flows through a solenoid, it establishes a magnetic flux.
Determination of the north and south poles of a solenoid
The solenoid has north polarity at the end where the magnetic lines of flux leave and a south polarity flux enter (return). The polarity depends on the direction of current flow and the direction of winding.
FORCE DETERMINATION
Force on a conduction carrying current in a magnetic field
Mechanical force exerted by the conclusion always acts in a direction perpendicular to the plane of the conductor and the magnetic field direction.
a) When a current carrying conductor is placed perpendicular (90o) to a magnetic field, reaction between the two magnetic fields is maximum
b) When a conductor is placed 0o to a magnetic field, there is no effect between them.
c) When a conductor is placed between 0o and 90o, only the perpendicular component is effective.
Force = Flux Density x current x length of conductor
F = BIL (N)
F = Force in conductor in Newton
B = Flux density in Tesla
I = current through the conductor in amperes
L = length of the conductor in meter
Numerical Examples
1) A conductor carries a current of 800A at right angle to a magnetic field having a density of 0.5T. Calculate the force on the conductor.
Data
Current (I) = 800A
Flux density (B) = 0.5T
Length (L) = 1m Force (F) =?
Force = flux density x current x length
F = BIL
= 0.5 x 800 x 1
= 400N
2) A conductor whose active length in a magnetic field is 0.5m carries a current 100A. If the flux density is 0.4T, calculate the force acting on the conductor.
Solution
Length (L) = 0.5m
Current (I) = 100A
Flux density (B) = 0.4T
Force (F) = ?
F = BIL
= 0.4 x 100 x 0.5
= 20N
3) A conductor 0.3m long is carrying a current of 60A at right angle to the magnetic field. The force on the conductor is 8N. Calculate the flux density of the magnetic field.
Solution
Length (L) = 0.3m
Current (I) = 60A
Force (F) = 8N
Flux density (B) = ?
F = BIL
B = 0.44T
ASSIGNMENT
1. A current – carrying conductor is situated at right angles to a uniform magnetic field having a density of 0.3T. calculate the force in Newton’s per meter length on the conductor when the current is 200A
2. A conductor, 150mm long is carrying a current of 60A at right angles to a magnetic field the force on the conductor is 3N. Calculate the density of the field.
3. A straight conductor carries a current of 5A perpendicular to a magnetic field of flux density 1.2T. If the force acting on the conductor is 2N, calculate the effective length of the conductor in a magnetic field.
Induced EMF due to motion
When a conductor rotates in a magnetic field, emf will be induced in the conductor. The induced emf can be calculated using the formula
E = BLVSinq, Volts
B is the flux density of the field (T)
E is the induced emf in the conductor in V
L is the length of the conductor (m)
V is the velocity (speed) at which the conductor is moving q is the angle at which the conductor lies in the magnetic field (degree)
Numerical Examples
1) A conductor of active length 0.3m moves in a magnetic field at a linear velocity of 500ms-1. If the magnetic flux density is 0.05T, calculate the value of the induced emf if the direction of movement of the conductor is
a. Parallel to the field
b. Perpendicular to the field
c. At 30o to the direction of the field.
Solution
Length (L) = 0.3m
Velocity (V) = 500ms-1
Flux density (B) = 0.05J
Emf =?
a. If parallel to the field
q = 0o = 0o
E = BLVsinq
= 0.05 x 0.3 x 500 x sin0
E = 0 V
b. If perpendicular to the field
q = 90o
E = 0.05 x 0.3 x 500 xsin90
E = 7.5V
c. If 30o to the field
q = 30o
E = 0.05 x 0.3 x 500 x sin30
E = 3.75V
2) A coil of active length 300mm rotates in a uniform magnetic field between the poles of an electromagnetic, which gives a field of flux density, 0.02T. if the conductor speed is 12.6m/s, determine
a. The maximum emf generated
b. The emf generated when the conductor is at 60o to the direction of the field.
Solution
Length (L) = 300mm = 0.3m
Flux density (B) = 0.02T
Speed (V) = 12.6ms-1
(a) E = BLV
= 0.02 x 0.3 x 12.6
= 0.0756V
(b) At 60o to the field
q = 60o
E = BLVsinq
= 0.02 x 0.3 x 12.3 x sin60
= 0.0655
DIRECTION OF INDUCED EMF
Two methods are available for deducing the direction of the induced or generated emf.
(1) Fleming‘s Right-Hand Rule
If the first finger of the right hand is pointed in the direction of the magnetic flux, and if the thumb is pointed in the direction of the motion of the conductor relative to the magnetic field, then the second finger, held at right angles to both the thumb and the second finger, represents the direction of the emf.
(2) Lenz’s Law
The direction of an induced emf is always such that it tends to set up a current opposing the motion or the change of flux responsible for inducing that emf.
In other words, the direction of the induced current must be that its own magnetic field will oppose the action that produced the induced current.
Induced EMF due to Flux charge
The laws of Electromagnetic Induction
Michael Faraday (1831)
Faraday’s Law
1. An induced emf is established in a circuit whenever the magnetic field linking that circuit is changed.
2. The magnitude of the induced emf is proportional to the ratio of change of the magnetic flux linking the circuit
From Faraday’s second law
Emf x rate of change of flux with time
E x Nϕ
t
E = Nϕt , dϕ = ϕ2 – ϕ1
p>
dt
Numerical Example
1. A magnetic flux 400µWb passing through a coil of 1200 turns is reversed in 0.1s. Calculate the average value of the emf induced in the coil.
Solution
Magnetic flux (f) = 400x10-6wb
Time (t) = 0.1s Emf = ?
Number of turns (n) = 1200t
E = Nϕt = 1200 𝑥 800𝑥10−6
dt 0.1
E = 9.6V
2. A magnetic flux 2mWb links with a coil of 100turns. If the flux
a) Doubled
b) Reversed in a time of 20ms. Calculate the value of the emf induced in the coil.
Solution
Magnetic flux (ϕ) = 2mw = 2x10-3wb
Number of turns (n) = 100turns
Time (t) = 2ms = 20x10-3sec.
(a) ϕ 1 = 2x10-3wb, double ϕ 2 = 4x10-3wb
E = Nϕt = 100 𝑥 (4-2)𝑥10−2
dt 20 x 10-2
E = 100 𝑥 (2) = 200
20
= 10V
(b) Reversed
ϕ 1 = 2x10-3wb ϕ 2 = 2x10-3Wb
E = Nϕt = 100 𝑥 (-2-2)𝑥10−2
dt 20 x 10-3
E = 100 𝑥 (-4) = -400
20 20
E = -20V
Self-Inductance
A circuit is said to have self-inductance when a change in its own current (by means of a variable resistor) cause a change in its magnetic flux thereby inducing emf into it.
The unit of self-inductance is Henry (H)
Henry (H)
Henry (H): A coil has self-inductance of 1hennry when a back emf of 1volts is induced in it. When the current through the coil is charging at the rate of 1ampere per second.
E = Ndϕ volts
dt
E = Ldϕ
volts
dt
Ndϕ = Ldϕ
dt dt
Ldt x dt = Ndϕ x dt
L = Ndϕ x dt
dt x dt
L = Ndϕ
dt
W = ½ LI2 Joules |
Numerical Examples
1) Calculate the average value of the emf induced in a coil of 0.4H inductance when in a time internal of 0.05seconds, the current
a. Increases from 0.1A to 2.1A
b. Reduces from 0.5A to 0.05A
Solution
Inductance (t) = 0.4H
Time (t) = 0.05sec
(a) I2 = 2.1A I1 = 0.1A
E = Ldϕ = L(I2 – I1)
dt t
= 0.4 𝑥 (2.1 – 0.1)
0.05
= 16V
b) I2 = 0.05A I1 = 0.5A
E = Ldϕ = L(I2 – I1)
dt t
= 0.4 𝑥 (0.05 – 0.5)
0.05
= -3.6V
2) A coil of 10turns produces a flux which changes from 0.5Wb to 2Wb and current 3A to 5A. Calculate the inductance of the coil.
Solution
Number of turns (N) = 10t
Flux (ϕ1) = 0.5wb
Flux (ϕ 2) = 2wb
Initial current (I1) = 3A
Final current (I2) = 5A
L = Ndϕ = N x (ϕ2 – ϕ1)
dt I2 – I1
= 100 𝑥 (2 – 0.5)
5 – 3
= 10 𝑥 (1.5)
2
L = 7.5H
3) A solenoid of self – inductance 5H carries a current of 2A. Determine the
a. Energy stored in the magnetic circuit if the current is reduced to 0.5A, determine the
b. New value of stored energy
c. Energy released into the electrical circuit
Solution
Induction (L) = 5H
Current (I) = 2A
(a) W = ½ LI2
= ½ x 5x (2)2
= ½ x 5 x 4
= 10J
(b) New current I = 0.5A
W = ½ LI2
= ½ x 5 x (0.5)2
= 0.625J
(c) Energy released = 10 – 0.625J
= 9.375J
Trial Test
1. A conductor of active length 0.5m moves in a magnetic field at a linear speed of 600ms-1. If the magnetic flux density is 0.04J, calculate the value of the induced emf of the direction of movement of the conductor is
a. At 45oto the direction of the field
b. Parallel to the field
c. Perpendicular to the field
2. State and explain two methods by which emf can be induced in a conductor.
3. State Faraday’s second law of electromagnetic induction
b. A magnetic flux of 300mWb passing through a coil of 1600turns is revered in 0.2sec.
Calculate the average value of the emf induced in the coil
4. A solenoid of self – inductance 6H carries a current of 3A. Determine the
a. Energy stored in the magnetic circuit If the current is reduced to 0.6A determine the;
b. New value of stored energy
c. Energy released into the electrical circuit
Eddy Current Loss
Eddy current loss is due to circulating currents set up in the core (steel lamination) of a machine.
It can be reduced by
a) Laminating the core
b) Using silicon – iron alloy. (the resistivity of silicon alloy is higher than steel)
Mutual Inductance
Two coils are said to be mutually coupled when a change in the magnetic flux produce by one coil causes an emf to be induced in the other coil. When majority of flux links both coils it is said to be mutually coupled. E.g. Power transformer. When small proportion of flux links both coils it’s said to be loosely coupled.
Measured in Henry (H)
E = MdI volts
dt
E = Ndϕ
dt
E = Ndϕ = MdI
dt dt
MdI x dt = Ndϕ x dI
MdI x dt = Ndϕ x dI
dI x dt dI x dt
M = Ndϕ
dI
Numerical Example
1) The mutual inductance between two coils is 0.15H. If the current in the primary winding increases from 0.2A to 0.5A in 10ms,
a. Determine the average value of emf induced in the secondary winding during this period of time
b. If the secondary winding is 400turns, determine the change in the flux which mutually links the coil
Solution
Mutual inductance (M) = 0.15H
Initial current (I1) = 0.2A
Final current (I2) = 0.05A
Time interval (t) = 10s – 0.01s
E = MdI volts
dt
dI = 0.5 – 0.2 = 0.3
= 0.15 𝑥 (0.5 – 0.2
0.01
= 4.5V
b. N = 400t
M = Ndϕ
dI
0.15 = 400 𝑥 dϕ
0.3
400dϕ = 0.15 x 0.3
dϕ = 0.15 𝑥 0.3
400
= 0.1125mwb
Application of Electromagnetism
1) Electric bell
2) Loud speaker
3) Solenoid
4) Buzzer
5) Measuring Instruments
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