G485 - Electric and Magnetic Fields

Question 1

How do electric field lines represent an electric field?

Answer 1

• arrows indicate the direction of the force that would act on a positive charge in the field
• density of the lines indicates field strength

Question 2

Coulomb’s Law

Answer 2

F = Qq / 4πε0r²

F = force of attraction or repulsion between two point charges Q and q

Question 3

What is the equation for electric field strength at any point in a radial field?

Answer 3

E = Q / 4πε0r²

Question 4

What is the equation for the force experienced by a charge in an electric field?

Answer 4

F = EQ

Question 5

What is the equation for electric field strength between charged parallel plates?

Answer 5

E = V/d

d = distance between plates

Question 6

Motion of Charged Particles in an Electric Field

Parallel Plates

Answer 6

Force acts at right angles to the original direction of motion
No effect on horizontal velocity
Particle follows a parabolic path

Question 7

Motion of Charged Particles in an Electric Field

Parallel Plates - Faster Speed

Answer 7

• more gradual curve

- longer path

Question 8

Motion of Charged Particles in an Electric Field

Parallel Plates - Slower Speed

Answer 8

• steeper curve

- shorter path

Question 9

Fleming’s Left Hand Rule

Answer 9

Thumb - F - Force
First Finger - B - Field
Second Finger - I - Current

Question 10

Right Hand Rule

Answer 10

Fingers - Current

Thumb - points in direction of North Pole

Question 11

Magnetic Flux Density

Answer 11

Amount of magnetic flux through a unit area taken perpendicular to the direction of the magnetic field

Question 12

Tesla

Answer 12

1 tesla is 1 Weber per metre squared

Question 13

What is the equation for the force acting on a current in a magnetic field?

Answer 13

F = BILsinθ

l = current
L = length of wire in the field
θ = angle between the field lines and the wire

Question 14

What is the equation for a charge moving perpendicular to a uniform magnetic field?

Answer 14

F = BQv

Question 15

Mass Spectrometer

Answer 15

Particle accelerated through a potential difference
Particle passes through a velocity selector
Particle passes into the detector

Question 16

Mass Spectrometer - Potential Difference

Answer 16

Particle is accelerated through a potential difference

W = VQ = 1/2 mv²

Question 17

Mass Spectrometer - Velocity Selector

Answer 17

Particle experiences a force in one direction from an electric field and in the opposite direction from a magnetic field
It can only pass out of the selector if it moves in a straight line, if the two forces are equal BQv = EQ
This is decided by velocity which is determined by mass when the particle accelerated through the p.d.

Question 18

Mass Spectrometer - Detector

Answer 18

The velocity of the particle when it enters the detector is equal to E / B1
Detector has a uniform magnetic field and the force exerted by that fields F = B2Qv
This force is also a centripetal force F = mv²/r
B2Qr = mE / B1
m = B1B2Qr / E
Radius is measured by the horizontal distance of the particle from the end of the velocity selector when it is detected

Question 19

Magnetic Flux

Answer 19

magnetic flux density x cross sectional area perpendicular to the field

Question 20

Weber

Answer 20

One weber is the magnetic flux when a magnetic field of flux density 1T passes at right angles through an area of 1m²

Question 21

What is the equation for magnetic flux?

Answer 21

ϕ = BA cosθ

Question 22

Magnetic Flux Linkage

Answer 22

Magnetic flux density x number of turns on the coil x area of coil perpendicular to field

Question 23

Faraday’s Law

Answer 23

E.M.F. Induced is proportional to the rate of change of flux linkage

E.M.F. = ∆Nϕ/ t

Question 24

Lenz’s Law

Answer 24

When a current is induced due to a magnetic field, the direction of the magnetic field around the induced current will oppose the direction of the original field

induced e.m.f. = -rate of change of flux linkage

Question 25

AC Generator

Answer 25

Coil rotated between the North and South Pole of a magnet
Slip rings maintain constant contact with the coil
Maximum change in flux linkage at 90 and 270 as the coil only has to rotate a small amount for a large change in the area perpendicular to the coil
Use root mean squared e.m.f. as the current is alternating so average e.m.f. would be 0

Question 26

Transformer - Turns Ratio

Answer 26

r = Vs/Vp = Ns/Np = Ip/Is

Question 27

Describe the Function of a Simple Transformer

Answer 27

• alternating emf applied to primary coil
• causing an alternating current in the primary coil which induced an alternating magnetic field
• alternating field induces an alternating field in the laminated iron core
• alternating field in the laminated core results in a change in flux linkage in the secondary coil
• by Faraday’s Law, an emf is induced on the secondary coil

Question 28

Why can’t DC be used for a transformer?

Answer 28

• there would be no change in flux linkage in the secondary coil and therefore no induced emf
• a transformer has to be used to reduce current so that the loss isn’t so great that the power received is too low to make anything work

Question 29

Transformers - Efficieny

Answer 29

Not 100% but can be up to 99% efficient

Question 30

Eddy Currents

Answer 30

• current induced in the core due to a change in flux linkage
• these currents have a magnetic field which opposes the field in the primary coil

-to reduce this the core is made up of laminated sheets rather than a solid block

Question 31

Inefficiencies in Transformers

Answer 31

Resistance of wires

Leakage inductance

Question 32

Electric Field Strength

Answer 32

Force per unit positive charge