Electric and Magnetic Fields

Question 1

What is a field?

Answer 1

A field is a region of space in which an object experiences a force

Question 2

What is an electric field?

Answer 2

An electric field is a region in space where the effect of an electrostatic force may be felt on a charge

Question 3

What is electric field strength?

Answer 3

Electric Field Strength = E = electric force F on a small charge/small charge Q

E = F/Q

Question 4

What does the electric field due to a point charge look like?

Answer 4

The direction of the field is the direction in which a small positive test charge would experience a force.

The electric field due to a point charge is radial. If the charge is positive, field lines away from it, if its negative then field lines into it. The closer to the charge, the stronger the field.

Question 5

What does the electric field look like due to:

a) two positive charges close to each other
b) a positive and a negative charge

Answer 5

a) Field lines don’t cross, neutral point in the middle

b) Field lines from the positive charge to the negative, like a magnet

Question 6

How do we show a uniform electric field?

Answer 6

To show a uniform electric field you can apply a p.d. to two metal plates dipped into a thin layer of insulating liquid. When short pieces of fine thread are sprinkled onto the liquid they line up in the shape of the field.

Question 7

What experiment do we use to demonstrate a uniform electric field?

Answer 7

To demonstrate that the electric field between two oppositely charged plates is constant, the following procedure can be used:

• Cut a test strip of aluminium foil, about 20mm by 5mm
• Attach it to the bottom of an uncharged insulating rod or plastic ruler
• Hold the ruler vertical and lower the end with its foil into the space between the two charged plates

The foil is charged by touching one of the plates. If the field is uniform, the force on the foil will be constant and it will hang at the same angle to the vertical; wherever it is placed in the field. If the charges on the plates are reversed, the test strip swings to the same angle the other way.

Question 8

What is Coulomb’s law?

Answer 8

Coulomb’s law: the electrostatic force between two point charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between them.

F = 1/4πEo * Q1Q2/r^2

Question 9

What force acts on two equal charges?

Answer 9

The electrostatic force of repulsion acts on two equal charges.
The forces actin on the charges are a Newton 3rd law pair.

The electrostatic force of repulsion can be increased by decreasing r(distance between charges) or increasing the charges.

Question 10

What is permittivity?

Answer 10

Permittivity is a measure of how electrostatic forces act through a medium. [F m^-1] = [A^2 s^4 kg^-1 m^-3]

Absolute permittivity(Eo) or the permittivity of free space is the permittivity in a vacuum, when two charges of 1C are placed 1m apart.

Relative permittivity(Er) is the factor by which the force between the charges is reduced (from the vacuum).

E (permittivity of medium) = Er * Eo

Question 11

What is the electric field strength due to a point charge?

Answer 11

F = kQ1Q2/r^2q => F/Q2 = kQ1/r^2

So radial field strength: E = kQ/r^2

Question 12

What experiment can we perform to investigate Coulomb’s law?

Answer 12

The forces are very small, so a sensitive measuring device is required to register the changes in force as the distance between the charges is changed.

Two conducting spheres are charged, by flicking the negative sphere with a woollen cloth and using the positive terminals of a d.c. supply set at about 30V to charge the upper positive sphere (or two positives or two negatives). Both charged spheres are insulated so they don’t use charge during the experiment.

The bottom charge is placed on a top pan balance on an insulating support, the top charge is hung above it, on an insulating rod.

To measure the distances between the spheres precisely, a light is shone onto the spheres and the distance can be measured from the shadows on the screen.

The electrostatic force acts on both spheres and hence the reading on the top pan balance changes with changes in distance.

Question 13

What is the electric field strength between two parallel plates?

Answer 13

The graph of potential difference over the distance between the plates is a straight downward sloping line.

E = V/d [N C^-1] [V m^-1]

Question 14

What is a capacitor?

Answer 14

Capacitor is a device which stores charge and energy. In its simplistic form a capacitor is just a pair of metal plates, separated by an insulator (dielectric)

Question 15

How can a capacitor be charged and discharged?

Answer 15

Charging a capacitor: electrons flow onto one of the plates and off the other one. The plates become oppositely charge. Charge can not pass directly from one plate to another because of the insulation between the plates.

The capacitor can be discharged by connecting it to a resistor (lamp, motor etc.). The discharge occurs very quickly, so the lamp lights briefly.

Question 16

What experiment can we perform to investigate how the charge stored on a capacitor varies with potential difference?

Answer 16

A circuit with a variable power supply, switch, voltmeter, a capacitor and two coulombmeters, either side of the capacitor, flying lead connected to one terminal of the capacitor.

1. Close the switch. The capacitor is joined to the power supply. Charge will flow onto the plates and the coulombmeter will give a reading.
2. Take the readings of p.d. and charge.
3. Open the switch
4. Discharge the capacitor by connecting the flying lead to the other terminal.
5. Repeat for more measurements.

The 2 coulombmeters give equal reading of charge, showing that the plates store equal and opposite charges.

Plotting a graph of charge over potential difference, it is a straight line through the origin.

Question 17

How does the charge stored by the capacitor vary with the potential difference across the plates?

Answer 17

From the experiment, the charge stored by the capacitor is directly proportional to the p.d. across the plates.

Q = k*V

Question 18

What is capacitance?

Answer 18

Capacitance is the charge stored divided by potential difference across the plates.

Question 19

What other property apart from capacitance do capacitors have?

Answer 19

Capacitors also have a working voltage. Exceeding this voltage causes the insulation between the plates to break down and conduct. This causes heating of insulation (usually explodes).

Question 20

What experiment can we perform to measure the working voltage of a parallel plate capacitor?

Answer 20

Using a bin liner as dielectric (insulator).

Connect an e.h.t. supply to two pieces of aluminium foil, put bin liner in between them (and under the bottom one). Gradually increase the p.d. until sparks start to appear on the bin liners. This means that they begin to conduct. Therefore this is the working voltage of this capacitor.

C = E*A/d => E = Cd/A (permittivity)

Question 21

How can we use a reed switch to measure the capacitance of a capacitor?

Answer 21

Use a circuit with a reed switch in the middle, a.c. supply and a diode connected to it, power supply on one side, with a voltmeter in parallel to it, resistor on the other side with an ammeter.

When the reed touches the left hand contact (power supplly) the capacitor is connected to the battery and so it charges up. When the reed is connected to the right hand side contact (resistor) the capacitor discharges through the ammeter.

Due to the a.c. supply, the direction of p.d. changes, but a diode is put in the circuit so that half-wave rectification occurs.

The voltmeter is connected to the power supply and not the capacitor because the capacitor discharges for half the time (the reading would be too small).

If the signal generator is set to f Hertz, the capacitor will charge and discharge f times per second. So f discharges of Q will pass through the microammeter each second.
I = fQ
I = CV*f (Q = CV)

Plot a graph of I over f, the gradient is CV.

At high frequencies the switch can not keep up with the changing magnetic field, so I is less than we expect.

Question 22

What is the energy stored when charging a capacitor?

Answer 22

By plotting a graph of V against Q, we can calculate the work done from the area. Hence:

W = 1/2 VmaxQ = VavQ
Using C = Qo/Vo
W = 1/2 VQ = 1/2 CV^2 = 1/2 Q^2/C (Electric Potential Energy)

Question 23

How can we measure the efficiency of a small electric motor with a capacitor?

Answer 23

Circuit: Power supply to a capacitor, switch , motor

Efficiency = MGH/1/2CV^2

Question 24

What are the formulae for total capacitance in

a) series
b) parallel

Answer 24

a) 1/Ct = 1/C1 + 1/C2 + 1/C3
b) Ct = C1 + C2 + C3

The opposite of resistance equations

Question 25

What are the formulae for discharge of a capacitor?

Answer 25

From the experiment where the a capacitor is discharge and time and charge/voltage/current are measured:

Q = Qo e^-t/CR
V = Vo e^-t/CR
I = -Io e^-t/CR

Question 26

What is the time constant?

Answer 26

CR is called the time constant (Tau = CR)

• measured in seconds as t/CR must have no units
• in a time of CR seconds we will have 37% of the starting
• 5CR is approximately 0.7% - practically discharged
• Half-life time: t=CRln2 => t (1/n) = CRln(n)

Question 27

What are the formulae for charging a capacitor?

Answer 27

Q = Qo(1-e^-t/CR)
V = Vo(1-e^-t/CR)
I = Io e^-t/CR

So the first two are growth to a maximum value (Qo, Vo), asymptote.
The third is exponential decay from maximum current.

Question 28

What is a magnetic field?

Answer 28

A magnetic field is a region in which magnetic materials (or moving electrical charges) feel a force

Question 29

What is magnetic flux density?

Answer 29

Magnetic flux density (B) is the magnetic field strength. It is measured in Teslas (T).

T = N A^-1 m^-1

Question 30

What is magnetic flux?

Answer 30

Magnetic flux (Ø) is a measurement of the total magnetic field which passes through a given area. It is measured in Webers (Wb)

Question 31

What is the formula for magnetic flux?

Answer 31

Ø = AB cos theta

Question 32

In which direction are magnetic field lines?

Answer 32

North to South

Question 33

How do we measure magnetic flux density?

Answer 33

With a magnetic flux density meter with a hall probe. A hall prove is a small piece of semiconductor layer. Four leads are connected to the midpoints of opposite sides. When control current is flowing through the semiconductor and magnetic field B is applied, the resultant Hall voltage Vh can be measured on the sides of the layer.
We must make sure that the field passes normally through the semiconductor slice, and that we measure the distance from the centre of the wires to the centre of the semiconductor.

Question 34

What do various magnetic fields look like?

Answer 34

The closer the field lines, the stronger the field.

In a magnet field lines go around from N to S.
When two magnets are close together there are two neutral points and the rest of the field is unchanged.

Electric currents also produce magnetic fields:

• use the right hand grip rule, thumb pointing in the direction of the current
• In a loop of wire with current, also use the right hand grip
• In a solenoid, looking into the solenoid S/N rule

Question 35

How does the field strength from a wire vary?

Answer 35

The greater the current, the higher the field strength (keeping distance constant)
The closer to the wire, the higher the field strength (keeping current constant)
B = k* I/r

B=µoI/2πr

µ is the permeability - it depends on the material around a wire

Question 36

How does the magnetic force on the wire vary with the current in the wire?(experiment + equation)

Answer 36

Put a manger on a top pan balance, lower the wire in between the magnet poles. Using Fleming’s left hand rule, the direction of the thrust on the wire (is a Newton’s 3rd law pair with the force on the top pan balance) can be determined.

The superposition of the circular magnetic field due to the current-carrying wire and the uniform field of the permanent magnet results in the field lines being compressed on the LHS of the wire (like a compressed spring) and a neutral point on the RHS (no resultant magnetic field).

From the experiment:
F is proportional to B, I and L

Therefore F = BIL

B has to be perpendicular to the direction of the current, so if it isn’t use F = BIL sin theta

Question 37

What is another expressions for a force on a wire, in terms of drift velocity?

Answer 37

The wire only experiences a force when electrons are moving in the wire (Flemings LH rule gives the direction of the force).
If the electrons are not moving in the wire there is no magnetic force.
B into the page, current moving from right to left
The individual electrons moving from left to right feel a force downwards. The total force downwards (BIL) is the sum of all the individual forces.
F=BIL (force on a wire)
F=BnAveL F/nAL = Bev

Question 38

What are Hemholtz coils?

Answer 38

Two identical coils, separated by the distance equal to their radius. The magnetic field strength everywhere inbetween the coils is uniform (constant field strength)

Question 39

What occurs during magnetic deflection of electrons?

Answer 39

The electrons are released during thermionic emission and then enter the magnetic field. The magnetic force (Bev) is always perpendicular to the velocity and therefore does not change the speed. Therefore the electrons travel in a circular motion.

F = ma
Bev = mv^2/r

1/2 mv^2 = eV

Use Pythagoras to find radius.

Question 40

How does an induced emf occur and how can it be increased?

Answer 40

When a wire cuts magnetic field lines we get an induced emf. It can be increased by:

1. Moving the wire with greater speed (v)
2. Increasing the magnetic field strength (B)
3. Using a longer length of wire (L)

E=BLv

Question 41

How is emf induced across a coil when a magnet is moved in and out of the coil? (Experiment)

Answer 41

When the magnet is moved into the coil it cuts the magnetic field lines. There is an induced emf across the terminals of the coil which we can measure using a voltmeter. We can see that:

1. If we use two magnets (B*2) the induced emf doubles
2. Moving the magnet with twice the speed doubles the emf
3. Using a coil with twice as many turns produces twice the induced emf (E proportional to N)

So, experiments show that the induced emf is proportional to the rate of change of magnetic flux (in the coil)
E proportional to dØ/dt

Question 42

What is Faraday’s law of electromagnetic induction?

Answer 42

Faraday’s law of electromagnetic induction: the induced emf in a circuit is equal to the rate of change of magnetic flux linkage through the circuit.

E proportional to d(nØ)/dt

Question 43

What is Lenz’s law?

Answer 43

When a magnet is pushed into a coil work must be done against a repulsive force if the current lights a bulb. The work done in moving the magnet is converted into electrical energy.

Lenz’s law: the direction of the induced emf is such that it tends to produce effects which oppose the change giving rise to it (ie the induced emf gives rise to an induced current in the coil which creates a magnetic field and therefore the magnet feels a magnetic force which opposes the motion of the magnet).

This is an application of conservation of energy.

Question 44

What is Neumann’s equation?

Answer 44

E = -d(NØ)/dt

Question 45

Describe an experiment where a magnet is dropped through a coil

Answer 45

The magnet will enter the coil with a smaller speed than it leaves the coil as the magnet is accelerating as it passes through the coil. t(in)>t(out)
The maximum induced emf when the current leaves the coil is bigger than the maximum emf when it enters the coil. This is because the rate of change of magnetic flux (∆Ø/∆t) is greater as the magnet leaves the coil (as it moves faster).
The induced emf must reverse as the magnet leaves the coil because of Lenz’s law.
Graph of emf over time. The areas under the curve are equal, the amplitude of the second peak is greater.

Question 46

Describe the experiment to investigate how the induced emf in the coil depends on the rate of change of flux linkage (Faraday’s law)

Answer 46

Two coils, one with a power supply and an ammeters the other with just a voltmeter. Stopwatch to measure time. We are going to induce an emf in coil 2 by having a changed magnetic flux.
The graph of induced emf in coil 2 against the rate of change of current in coil 1 is a straight line passing through the origin.
So E is proportional to dI/dt; B is proportional to I, so E is proportional to dB/dt; Ø=AB, therefore E is proportional to dØ/dt, induced emf is proportional to rate of change of flux linkage.

Question 47

Derive BLv from conservation of energy

Answer 47

Power = Work Done/Time
EI = BIL*v/1
Electrical power = Mechanical power
E =Blv

Question 48

Derive Blv from Neumann’s equation

Answer 48

E = -N dØ/dt
E = -NB dA/dt
In one second the wire moves v metres. The length of the wire is L, so the area swept out per second is Lv => dA/dt = lv/1
E = -B(lv/1)
Therefore E = -Blv

Question 49

Describe how an AC generator works

Answer 49

A coil turns in a constant magnetic field with constant angular velocity
Ø = ABcosø
NØ = NABcosø

Question 50

What is an eddy current?

Answer 50

Eddy currents - large circulating currents in conductors that produce heating of the conductor
We can use Fleming’s RHR to find the direction of the induced current. The sheet feels a repulsive force as it enters the field. The KE of the sheet is transformed into thermal energy in the aluminium.
This is electromagnetic braking. The induced emf can be used to charge batteries in cars and buses. If the sheet has strips then large eddy currents can not flow so effect of damping of the oscillations is greatly reduced.