Electromagnetic induction

Question 1

What happens when a conducting rod is moved perpendicularly through a magnetic field?

Answer 1

• The magnetic field will exert a force on the ‘free’ electrons in the rod, causing them to move [downwards]. Electrons accumulate at the top/bottom] of the rod, and an electric field is created between the two ends of the rod. Electrons stop moving once there is enough to repel new electrons.

Question 2

How can you calculate the electric field created between the ends of a rod moving normally to a magnetic field?

Answer 2

• E = Ɛ/L
• Induced emf, Ɛ is the p.d. between the ends of the rod.
• L is the length of the rod

Question 3

What is motional emf? How can you calculate it?

Answer 3

• The p.d. between the ends of a conducting rod length L moving at speed v normally to a magnetic field B.
• Ɛ = BvL

Question 4

What are the ways to increase the current induced by moving a coil of wire through a magnetic field? [5]

Answer 4

• Increase the relative speed of the magnet and the coil
• Increase the strength of the magnet
• Increase the number of turns in the coil
• Increase the area of the loop
• Move the magnet at right angles to the plane of the loop.

Question 5

What is magnetic flux? How can you calculate it?

Answer 5

Imagined as the number of field lines piercing a loop of wire in a magnetic field. Called flux linkage when there is more than one turn of wire.
- Φ = NBAcosθ (Weber - Wb)
N = Number of turns in coil
B = magnetic field strength
A = Area of the loop in the magnetic field.
θ = angle between the magnetic field and the normal to the loop area

Question 6

What is Faraday’s law?

Answer 6

The induced emf (not necessarily current) of a coil in a magnetic field is equal to the negative rate of change of magnetic flux linkage.
- Ɛ = - NΔΦ/Δt

Question 7

How can you work out motional emf of a rod in a magnetic field using Faraday’s law?

Answer 7

• Imagine the area swept out by the rod as L x vΔt

- Substitute this into the equation for change in magnetic flux. ΔΦ = BA

Question 8

What is Lenz’s law?

Answer 8

• the induced emf will be in such a direction as to oppose the change in magnetic flux that created the current.
• (the magnetic field produced by the induced current must be in the opposite direction to the magnetic field creating the induced current - use RHG rule)

Question 9

How does an ac generator work?

Answer 9

The ends of the coil are attached to slip rings which are in contact with carbon brushes. The flux in the coil changes as the coil rotates and so an emf is produced in it.

Question 10

How can you calculate and graph the induced emf in an ac generator?

Answer 10

V = -ΔΦ/Δt = ωNBAsin(ωt)

Imagine a sine graph of voltage against time.

Question 11

How can you calculate and graph the current produced by an ac generator?

Answer 11

• I = V₀sin(ωt)/R
  V₀ is the peak voltage
  ω is the angular speed of the coil
  Imagine a sine graph of current against time

Question 12

How can you calculate the power in an ac circuit?

Answer 12

P = V₀I₀sin²(ωt)
V₀ is the peak voltage
I₀ is the peak current
ω is the angular speed of rotation of the coil
∴ Pmax = V₀I₀
- Two oscillations on the graph corresponds to one rotation of the coil.

Question 13

How can you find the average voltage, current and power in an ac circuit?

Answer 13

Use the root mean square (rms):
Irms = I₀/(√ 2)
Vrms = V₀/(√ 2)
P[bar] = V₀I₀/2 = IrmsVrms

Question 14

How can ac current be converted to dc current?

Answer 14

• Using a single diode (half wave rectification. Graph turns into a series of flats and hills. Has the disadvantage that half the power is lost.
• Using a diamond of diodes all pointing in the sameish direction, with two circuits connected to opposite corners of the diamond. Graph is a series of hills.

Question 15

What is capacitance?

Answer 15

The charge per unit voltage that can be stored on a capacitor (a property of the capacitor): C = q/V = εA/d (F)
q = the charge on one of the plates
V is the potential difference between the plates
ε is the permittivity of the medium between the plates
A is the area of overlap
d is the plate separation

Question 16

What is a dielectric? How does it work?

Answer 16

• It is an insulator inserted between the plates of a capacitor to increase its capacitance.
• The electric field created between the plates of the capacitor exerts a force on the charges of the dielectric, causing charge polarization. This creates a small electric field within the dielectric, reducing the net electric field between the plates.

Question 17

How can you work out the total capacitance of two or more capacitors connected in parallel?

Answer 17

Add the individual capacitors together:

Cparallel = C₁ + C₂ …

Question 18

How can you work out the total capacitance of two or more capacitors in series?

Answer 18

1/Cseries = 1/C₁ +1/C₂ …

Question 19

How can you calculate the energy stored in the electric field between the plates of a capacitor?

Answer 19

Find the area under a p.d. vs charge stored graph.
E = (1/2)CV²
E is the energy stored
C is the capacitance
V is the p.d. between the plates
Substitute quantities using C = q/V for more equations.

Question 20

What is a circuit set-up for charging a capacitor?

Answer 20

Use a two way switch that can connect the capacitor to two circuits.

Question 21

What does the graph look like for a charging capacitor?

• p.d or charge against time
• current against time

Answer 21

• A curve passing through the origin that gradually flattens off. The p.d. or charge increases until V = Ɛ emf of the battery and q = CƐ
• A curve that tends to zero, starting at I₀ = Ɛ/R

Question 22

What is the time constant (for a capacitor)?

Answer 22

τ = RC = resistance x capacitance
The time constant is the time after which the charge on a discharging capacitor decreases to about 37% of its initial value.

Question 23

the time constant is the time needed for the charge to decrease to half its initial value, divided by ln 2

Answer 23

aPAGE 2 STILL TO DO

Question 24

How can you smoothen the rectified output dc current from an ac input?

Answer 24

Put a capacitor in parallel to the load. As the voltage increases, the capacitor stores charge. As the voltage decreases, the capacitor discharges, sending current through the load, and ‘smoothening/flattening’ the voltage time graph.

Question 25

How does a transformer work?

Answer 25

A primary coil carries an alternating current around a soft iron core. The changing current creates a changing magnetic field inside the core. Because the magnetic field is changing, the flux is changing, inducing a current in the secondary coil. Hence it does not work with dc current.

Question 26

How can you calculate the current, voltage, and frequency in a transformer? How can you derive the equation for current?

Answer 26

• Frequency is the same in both coils
• Ɛp/Ɛs = Np/Ns = Is/Ip DB
  Assuming it is an ideal transformer (power in = power out), ƐpIp = ƐsIs, rearrange.

Question 27

What are some ways that energy is lost in a transformer? [2]

Answer 27

• Resistance heating in the coils: Ploss = RI²

- Eddy current heating in the iron core, reduced by laminating the core.

Question 28

What is the:

• work done moving an electron from the top to bottom of a moving rod in a magnetic field?
• Force on the electron?
• Work done per unit charge?

Answer 28

• W = FL = evBL (rod length L)
• F = evB (v is the rod speed)
• Work per unit charge = emf Ɛ = BvL

Question 29

What is magnetic flux density?

Answer 29

The same as magnetic field strength, B