Electromagnetism Flashcards

Magnetism

Magnetism is a phenomenon where materials exert attractive or repulsive forces on each other. Iron, nickel, cobalt, and their alloys exhibit magnetic properties due to the motion of electrons. Moving electric charges create magnetic fields. Every magnet has a north and south pole, which cannot be isolated.

Magnetic Fields

A magnetic field is the region around a magnet where it has an effect. Magnetic field lines indicate the direction a ‘magnetic north’ would travel. To plot the magnetic field due to a bar magnet, you need a bar magnet, plotting compasses, and a sheet of cardboard. Place the bar magnet on the paper, position the compass, and mark the paper at either end of the needle. Repeat the process to trace the field lines.

STS: Earth’s Magnetic Field

The Earth has a magnetic field, believed to be due to convection currents of molten iron in the Earth’s core. It is used for navigation and protects us from the solar wind by deflecting charged particles toward the poles, creating the Northern and Southern Lights. This radiation allowed evolution to occur on Earth.

Magnetic Field due to Current

In 1819, Hans Christian Oersted observed that a compass needle points tangent to circles radiating from a current-carrying conductor. The direction of the field can be found using the ‘right hand grip rule’ or ‘Maxwell’s corkscrew’ rule. For a long straight wire, the field is a series of circles. For a loop, the field lines form circles around the loop. A solenoid's field looks like a bar magnet. A solenoid is a coil of insulated wire, cylindrical in shape and with a length much greater than its diameter.

Demonstration: Magnetic Field

To plot the magnetic field, you need a battery, leads, switch, plotting compasses, wire, loop, solenoid, and cardboard. Set up circuits, close the switch, and use compasses to map the magnetic field.

STS: Electromagnets and Their Uses

An electromagnet is a magnet constructed from a soft iron core with a coil of wire. The magnetic field appears when current passes through the wire. They are used in scrap yards, electromagnetic relays, MCBs, loudspeakers, tape recorders, and maglev trains.

Magnetic Flux Density

Magnetic flux density (B) measures the strength of the magnetic field. It is a vector quantity with units of tesla (T). The direction is along field lines from north to south. A magnetic field of 1 tesla exerts a force of 1 newton on a 1-ampere current in a 1-meter conductor within the field: 1 T = 1 N A^{-1} m^{-1}.

Force on a Current-Carrying Conductor in a Magnetic Field

When the current is perpendicular to the magnetic field, the force (F) is proportional to the current (I), length (l), and magnetic field strength (B): F = IlB. The more accurate formula is F = IlB[Theta], where [Theta] is the angle between the current and the field. If the conductor is parallel to the field ([Theta] = 0), there is no force.

Fleming’s Left-Hand Rule

Use your left hand, pointing your thumb, first finger, and second finger at right angles. The thumb indicates the force (F), the first finger represents the magnetic flux density (B), and the second finger represents the current (I). Remembering FBI may help.

Demonstration of Force

To demonstrate force, use aluminium foil, a battery, switch, resistor, ammeter, and leads. The foil moves perpendicular to the current and magnetic field. Reversing the current changes the direction of movement. The force depends on length (l) and current (I).

If you plot force (F) against current (I), the slope (m) of the graph would be lB. The magnetic flux density can be calculated as B = \frac{m}{l}.

Sample Problem EB3 A
A current of 5 A flows through a straight piece of wire that passes through 4 m of uniform magnetic field of flux density 1.5 T, at right angles to the direction of the field. What is the magnitude of the force experienced by the wire?

Sample Answer EB3 A
Using the formula for force on a current-carrying conductor in a magnetic field:
F = IlB = (5)(4)(1.5) = 30
The wire experiences a force of 30 N.

Sample Problem EB3 B
A rectangular loop of wire, carrying a current of 0.5 A, is placed in a uniform magnetic field of flux density 6 T. The loop of wire, with dimensions 30 cm × 10 cm, is free to rotate about a central axis. (i) Calculate the magnitude of the force acting on each of the 30 cm sides of the loop. (ii) Calculate the torque of the couple about the central axis when it is in the position shown. (iii) Describe what happens to the torque of the couple as the loop rotates through 90˚. (iv) Will the loop of the wire stop moving when it has rotated through 90˚? Explain. (v) What would happen if the direction of the current changed as the loop rotated through 90˚?

Sample Answer EB3 B
(i) F = IlB = (0.5)(0.3)(6) = 0.9 N
(ii) T = Fd = (0.9)(0.1) = 0.09 Nm
(iii) The torque decreases to zero as the loop rotates because the perpendicular distance decreases.
(iv) The loop will continue moving due to its momentum.
(v) The rotation would continue if the current direction changed.

STS: Applications of Current in a Magnetic Field

The simple d.c. motor is based on the principles of Sample Problem EB3 B, as is the moving-coil galvanometer and the moving-coil loudspeaker.

Force on a Charged Particle Moving Through a Magnetic Field

Current is the flow of electric charge. Even a single charge moving constitutes a current and experiences a force in a magnetic field. Fleming’s left-hand rule is used to find the force direction. The direction of (conventional) current is from positive to negative. The force on the particle is F = qvB, where q is charge, v is velocity, and B is the magnetic flux density.

Derivation:
F = IlB
F = \frac{QlB}{t},
F = QvB.
Since Q = nq and n = 1, then F = qvB.

If a charged particle enters a magnetic field perpendicularly at a constant speed, it moves in a circle. The centripetal force is F = \frac{mv^2}{r}.

Sample Problem EB3 C
An electron moving at 5 × 10^4 m/s moves at right angles to a uniform magnetic field of flux density 7 T. Given that an electron has a charge of 1.6 × 10^{-19} C, what is the force on the electron?

Sample Answer EB3 C
F = qvB = (1.6 × 10^{-19})(5 × 10^4)(7) = 5.6 × 10^{-14} N

Forces Between Currents

Two current-carrying conductors exert forces on each other. Parallel wires with current flowing in the same direction attract, and those with current in opposite directions repel.

Definition of the Ampere

The ampere is the constant current which, if maintained in two straight parallel conductors of infinite length, of negligible cross section and placed 1 metre apart in a vacuum, would produce a force on each conductor of 2 × 10^{-7} newtons per metre of length.
Demonstration:
To demonstrate forces, use two strips of aluminium foil, a battery, switch, resistor, ammeter, and leads. The strips move away from each other, demonstrating the force between current-carrying conductors due to their magnetic fields.

Electromagnetic Induction

Electromagnetic induction is where an emf is induced by a change in the magnetic flux linking a circuit.

Magnetic Flux

Magnetic flux (\Phi) is the total amount of magnetic field in a region: \Phi = BA, where B is magnetic flux density and A is area. The SI unit is the weber (Wb). 1 Wb = 1 T m^2.

Faraday’s Law

The magnitude of the induced emf is proportional to the rate of change of flux linking a circuit: E = \frac{d\Phi}{dt}. For numerical calculations:

induced emf = \frac{final \, flux - initial \, flux}{time \, taken}.

Sample Problem EB3 D
A loop of wire of area 0.1 m^2 is rotated within a magnetic field of flux density 9 T, completing 20 revolutions per minute. Calculate the size of the induced emf.

Sample Answer EB3 D
Time for each revolution: T = \frac{60}{20} = 3 s. The flux changes from maximum to zero in 0.75 seconds.
\Phi = BA = (9)(0.1) = 0.9 Wb.
induced emf = \frac{0.9 - 0}{0.75} = 1.2 V.

Demonstration of Faraday’s Law

To show induction, use a coil, galvanometer, and magnet. Moving the magnet towards or away from the coil induces current, as indicated by the galvanometer. Faster movement yields a larger reading.

Lenz’s Law

The direction of the induced emf opposes the change causing it.

Demonstration of Lenz’s Law

Using a copper pipe, time how long the magnet takes to fall compared to a non-magnetised metal. The magnet takes longer due to the opposing force from induced eddy currents.

Demonstration of Lenz’s Law with Aluminium Ring

Pushing the magnet towards the ring causes it to move away, and pulling it away causes the ring to move toward it. This induction creates an opposing magnetic field in the ring.

Change of Mechanical Energy to Electrical Energy

Lenz’s law ensures that the law of conservation of energy holds true. When a magnet approaches a loop of wire, the induced emf's electrical energy comes from the work done in moving the magnet.

STS: Electromagnetic Damping

Damping is the slowing of a body's vibration or movement. Electromagnetic damping has no physical contact and thus no wear. The energy conversions are kinetic to electrical to heat.

Demonstration of Electromagnetic Damping

Spin an aluminium disc and bring a magnet close. The disc stops spinning due to induced eddy currents opposing the rotation.

By combining Lenz’s law with Faraday’s Law we get: E = -\frac{d\Phi}{dt}. In a solenoid: E = -N\frac{d\Phi}{dt}

Sample Problem EB3 E
A rectangular planar coil of wire with 200 turns and dimensions 3 cm by 4 cm is rotated at constant speed of 8\pi rad s^{-1} within a uniform magnetic field of flux density 2.5 T. Given that the axis of rotation is perpendicular to the magnetic field, calculate the induced emf in the coil.

Sample Answer EB3 E
Using the formula for periodic time of circular motion, area of a rectangle, magnetic flus and the induced emf in a solenoid, making sure all values are in their standard unit…
Time taken for each revolution is…
T = \frac{1}{5} = 0.2 \, s
Area = length × width ⟹ A = (0.03)(0.04) = 1.2 × 10^{-3} \, m^2
When the magnetic field passes through the ‘full area’ the total flux is given by: \Phi = BA = (2.5)(1.2 × 10^{-3}) = 3 × 10^{-3} \, Wb.
Time taken for flux to change from 3 × 10^{-3} \, Wb to 0 Wb every 0.025 s.
E = -N \frac{d\Phi}{dt} = -(200) \frac{0.003}{0.025} = -24
The induced emf is 24 V.

STS: Applications of Electromagnetic Induction in Generators

Generators convert mechanical energy into electrical energy by rotating a coil of wire in a magnetic field, inducing an emf. This induced emf opposes the rotation (Lenz’s Law). Most generators produce an alternating voltage.

Alternating Current

An alternating current (a.c.) periodically reverses direction. Plotting voltage against time yields a sinusoidal graph. A direct current (d.c.) flows continuously in the same direction, typically from a battery.

Demonstration of AC and DC

Connect a signal generator to an oscilloscope. Alternating voltage produces a sinusoidal pattern, while constant voltage produces a straight line.

STS: National Grid and AC

Mains electricity is a.c. because it is easier to generate and more efficient to transport. It can be stepped up to high voltage using transformers, reducing energy losses due to heat. Lower current allows for thinner, less expensive wires. Voltages are stepped down for local use (e.g., 230 V for homes).

Peak and RMS Values

The peak voltage (V0) is the maximum voltage reached. The root mean squared voltage (V{rms}) is equivalent to a d.c. supply.
Alternating Voltage
V{rms} = \frac{V0}{\sqrt{2}}

Alternating Current
I{rms} = \frac{I0}{\sqrt{2}}

Sample Problem EB3 F

The rms voltage of the mains electricity is 230 V. Calculate the peak voltage of the mains?

Sample Answer EB3 F

V{rms} = \frac{V0}{\sqrt{2}} \implies 230 = \frac{V0}{\sqrt{2}} \implies V0 = 230 \sqrt{2} = 325
The peak voltage of the mains is 325 V.

Mutual Induction

Mutual induction occurs when a changing magnetic field in one coil induces an emf in another coil. The size of the induced emf depends on the ratio of turns in each coil.

Demonstration of Mutual Induction

Use two coils, a battery, switch, galvanometer, and leads. Closing the switch in circuit A (primary) induces a current in circuit B (secondary) when the two circuits close to each other, demonstrating mutual induction.

Transformers

A transformer changes the voltage of a.c. power supplies using mutual induction. It consists of a primary coil and a secondary coil wrapped around a soft iron core.

Demonstration of a Transformer

Use a transformer, battery, switch, two filament lamps, and two voltmeters. The ratio of turns in the primary and secondary coils determines whether the output voltage is higher or lower than the input.

Calculating Output Voltage

The ratio of turns is equal to the ratio of voltages:
\frac{Vi}{Vo} = \frac{Np}{Ns}

Transformer Efficiency

Transformers are less than 100% efficient due to energy losses, typically as heat. Laminating the core reduces eddy currents and improves efficiency. Factors affecting efficiency: eddy currents, heat losses, core shape, core laminations.

STS: Uses of Transformers

• Mobile phone chargers use step-down transformers to recharge low-voltage batteries.
• Televisions use step-down and step-up transformers.
• Computers use step-down transformers.
• Power stations use step-up and step-down transformers in the national grid.

Sample Problem EB3 G

Mains electricity has a voltage of 230 V. Using a transformer with 3600 turns in the primary coil and 180 turns in the secondary, what will the output voltage be?

Sample Answer EB3 G

\frac{Vi}{Vo} = \frac{Np}{Ns} \implies \frac{230}{Vo} = \frac{3600}{180} \implies Vo = \frac{230 × 180}{3600} = 11.5
The output voltage is 11.5 V.

Self-Induction

Self-induction is where a changing magnetic field in a coil induces an emf in the coil itself. The induced emf (back emf) opposes the voltage across the coil. When the current stops, the back emf is much greater. When an a.c. voltage is applied across a coil, there is a constantly changing magnetic field.

Inductors

An inductor induces a back emf and reduces current in a circuit with a constant a.c. supply. It offers resistance based on its resistivity, length, and cross-sectional area. Inductors do not generate back emf when a battery drives a direct current.

Demonstration of Self-Induction

Use a coil, battery, two switches, and a neon lamp. Closing switch A does not light the bulb, but opening it causes a flash due to the collapsing magnetic field.

STS: Uses of Inductors

• Dimmer switches vary inductance to control current and light brightness.
• Inductors are used in radios with capacitors to select station frequencies.