7. Electric and Magnetic Fields

Question 1

What is a force field?

Answer 1

An area in which an object experiences a non-contact force.

Question 2

How can force fields be represented?

Answer 2

• Vectors.

- Field lines.

Question 3

What is an electric field?

Answer 3

A force field in which a charged particle experiences a force.

Question 4

What is the electric field strength?

Answer 4

The force per unit charge experienced by an object in an electric field.

Given by the formula E = F/Q

Question 5

What is the difference between the electric field strength of uniform fields and radial fields?

Answer 5

In uniform fields the value of the electric field strength is constant.

In a radial field, it varies.

Question 6

What does coulombs law state?

F = (Q₁Q₂) / 4πε₀r²

Answer 6

The magnitude of the force between two point charges is:

• directly proportional to the product of their charges.
• inversely proportional to the square of the distance between them.

Question 7

what is ε₀ and r in:

F = (Q₁Q₂) / 4πε₀r²

Answer 7

ε₀ is the permittivity of free space.

r is the distance between charges (from their centres).

Question 8

What happens if F is negative in coulombs law?

F = (Q₁Q₂) / 4πε₀r²

Answer 8

The force is attractive if F is negative.

The fore is repulsive if F is positive.

Question 9

When do you use this equation?

E = Q/4πε₀r²

Answer 9

To find the electric field strength of a point charge forming a radial field.

Question 10

What is the absolute electric potential (V)?

Answer 10

The potential energy per unit charge of a positive point charge at a point in the field.

Question 11

How does the absolute magnitude of electric potential change in a radial field?

Answer 11

• It is greatest at the surface of the charge.
• as the distance from the charge increases, its potential decreases.
• Its electric potential at infinity is 0.

Question 12

What happens to the potential and polarity if the sign of the charge is positive/negative?

Answer 12

If a charge is positive…

• its potential is positive.
• its charge is repulsive.

If a charge is negative…

• its potential is negative.
• its charge is attractive.

Question 13

What does V-r look like for a repulsive charge?

What does V-r look like for an attractive charge?

(V = electric potential)
(r = distance from centre of charge)

Answer 13

Repulsive:

• As V decreases, r increases at a growing rate.
• V is initially positive.

Attractive:

• As V increases, r increases at a growing rate.
• V is initially negative.

Question 14

When do you use this equation?

E = V/d

Answer 14

To find the electric field strength in a uniform electric field.

Question 15

How can you form a uniform electric field?

Answer 15

Using parallel plates with a potential difference across them.

Question 16

What is electric potential difference?

Answer 16

The energy needed to move a unit charge between two points.

Question 17

What do field lines show?

in an electric field

Answer 17

The direction of the force acting on a positive charge.

Question 18

What does the distance between field lines represent?

Answer 18

The magnitude of force.

Question 19

What do equipotential lines show?

Answer 19

The points where the potential in a field are the same.

• When a charge moves along an equipotential line, no work is done.

Question 20

What is capacitance?

Answer 20

The charge stored by a capacitor per unit potential difference.

Question 21

What is the area under a graph of charge against potential difference for a capacitor?

Answer 21

The electrical energy stored by a capacitor (W).

Question 22

Describe the graph of charge against potential difference for a capacitor?

Answer 22

A straight line through the origin.

as potential difference is directly proportional to charge

Question 23

When charging a capacitor, what do the graph of:

I-t
V-t
Q-t

…look like?

Answer 23

I-t
decreasing slope
(are under = charge)

V-t
Increasing slope

Q-t
Increaing slope
(gradient =current)

Question 24

describe what happens when you charge a capacitor to explain the graphs of I-t, V-t and Q-t?

Answer 24

Once a capacitor is connected to a power supply…

• current starts to flow and negative charge builds up on the plate connected to the negative terminal.
• On the opposite plate, electrons are repelled by the negative charge building upon the initial plate.
• Therefore electrons move from the positive terminal and an equal but opposite charge is formed on each plate.
• This creates a potential difference across the plate.
• As the charge across the plate increases, the potential difference increases but the electron flow decreases.
• This is due to the electrostatic repulsion also increasing.
• Therefore the current decreases — eventually to zero.

Question 25

When discharging a capacitor, what do the graph of:

I-t
V-t
Q-t

…look like?

Answer 25

I-t
decreasing slope
(are under = charge)

V-t
decreasing slope

Q-t
decreasing slope
(gradient =current)

Question 26

Why are the discharging graphs of I-t, V-t, Q-t the way they are?

Answer 26

When the capacitor is dischargin…

• the current flows in the opposite direction.
• and the current charge and potential difference across the capacitor falls exponentially
• They take the same amount of time for each value to half.

Question 27

What is the time constant when talking about capacitors?

Answer 27

• RC
• it is the time taken for it to discharge to 1/e (0.37) of its initial value (of charge, current and voltage)
• it is the time taken for it to charge to 1 - 1/e (0.63) of its maximum value (of charge or voltage)

Question 28

How do you find the time constant for a capacitor from a graph?

Answer 28

IF the graph is decreasing, find the time at y = 0.37

ELSE IF the graph is increasing, find the time at y = 0.63

Question 29

How would you plot Q = Q₀e⁻ᵗ/ᴿᶜ in the form y = mx + c

Answer 29

Q = Q₀e⁻ᵗ/ᴿᶜ

ln(Q) = ln(Q₀e⁻ᵗ/ᴿᶜ)
ln(Q) = ln(Q₀) + ln(e⁻ᵗ/ᴿᶜ)
ln(Q) = ln(Q₀) - t/RC

ln(Q) = -1/RC(t) + ln(Q₀)
y=m(x) + C

plot ln(Q) angaint t

Question 30

Derive V = V₀e⁻ᵗ/ᴿᶜ and I = I₀e⁻ᵗ/ᴿᶜ from Q = Q₀e⁻ᵗ/ᴿᶜ.

Answer 30

Q = Q₀e⁻ᵗ/ᴿᶜ
C = Q/V ∴ Q=CV

Q₀=CV₀
CV = CV₀e⁻ᵗ/ᴿᶜ
V = V₀e⁻ᵗ/ᴿᶜ

V=IR
V₀=I₀R

IR = I₀Re⁻ᵗ/ᴿᶜ
I = I₀e⁻ᵗ/ᴿᶜ

Question 31

What is the magnetic flux density of a magnetic field?

Answer 31

• A measure of the strength of the field.

- Measured in Tesla.

Question 32

What is magnetic flux?

Answer 32

Symbol: Φ

describes the magnetic field or magnetic field lines passing through a given area.

Question 33

How do you work out magnetic flux when the field is perpendicular to the area?

Answer 33

Φ = BA

B is the magnetic flux density
A is the area

Question 34

What is magnetic flux linkage?

Answer 34

Symbol: NΦ

it’s the magnetic flux multiplied by the number of turns.

Question 35

Why does a current-carrying wire experience a force in a magnetic field?

Answer 35

• it has electrons that are charged and so a force acts of them in a magnetic field.
• A particle has a force of magnitude BQvSinθ
• The wire has a force of magnitude BIlSinθ.

Question 36

Describe electromagnetic induction?

Answer 36

• A conducting rod/bar magnet moves relative to a magnetic field/a coil of wire.
• It experiences a rate of change of flux linkage.
• This causes an equal magnitude of emf to be induced in the rod. (Faraday’s Law)
• If the coil of wire forms a complete circuit, a current is induced which opposes the motion causing it.(Lenz’s Law)

Question 37

What factors affect the emf induced in a coil during electromagnetic induction?

Answer 37

• The number of turns: directly proportional (due to Faraday’s law).

- The magnetic flux density:
directly proportional (as Φ = BA).

• The cross-sectional area of coil: directly proportional (as Φ = BA).
• Time taken for motion: inversely proportional (due to Faraday’s law).

Question 38

Describe electromagnetic induction in transformers?

Answer 38

• The primary coil carries an alternating current.
• The current-carrying wire creates a changing magnetic field in the iron core.
• The rate of change of flux linkage across the second wire induces an emf in the coil. (due to Faraday’s law).
• because the coil is in a closed circuit a current is also induced. This acts in the opposing direction to the current in the primary coil (due to Lenz’s law).

Question 39

What factors affect the emf induced in the secondary coil of a transformer?

Answer 39

The magnetic flux density created by the initial coil:

• directly proportional.
• determined by the number of turns (l) and the current (I) flowing through the wire.

The distance between the two coils:

• inversely proportional.
• as the further they are, the less magnetic flux passes through the second coil and so the induced emf is lower.

The number of turns in the second coil:
- directly proportional.

The cross-sectional area of the second coil:
- directly proportional.

The frequency of alternating current:

• directly proportional.
• dependent on the time take for the current to change.

Question 40

How does a magnet falling through a coil demonstrate Lenz’s law?

Answer 40

• As the magnet approaches the coil, there is a rate of change of flux linkage through the coil, so an emf and currents are induced.
• The direction of the current opposes the motion of the magnet, so the same pole as the pole the magnet approaching the coil will be induced in order to repel it.
• This causes the magnet to slow down due to the force of repulsion.
• As the magnet passes through the centre of the coil there is no change in flux linkage so no emf is induced.
• As the magnet leaves the coil the is a rate of change of flux linkage, so en emf and current are induced that opposes the motion of the magnet.
• Thus an opposing pole is induced by the agent causing it to slow down once again, due to the force of attraction.

Question 41

Why is UK voltage given as 230V if it peaks at 330V?

Answer 41

230V is the root mean squared.

330/√2 ≈ 230V