G485 - Electric and Magnetic Fields Flashcards
How do electric field lines represent an electric field?
- arrows indicate the direction of the force that would act on a positive charge in the field
- density of the lines indicates field strength
Coulomb’s Law
F = Qq / 4πε0r²
F = force of attraction or repulsion between two point charges Q and q
What is the equation for electric field strength at any point in a radial field?
E = Q / 4πε0r²
What is the equation for the force experienced by a charge in an electric field?
F = EQ
What is the equation for electric field strength between charged parallel plates?
E = V/d
d = distance between plates
Motion of Charged Particles in an Electric Field
Parallel Plates
Force acts at right angles to the original direction of motion
No effect on horizontal velocity
Particle follows a parabolic path
Motion of Charged Particles in an Electric Field
Parallel Plates - Faster Speed
- more gradual curve
- longer path
Motion of Charged Particles in an Electric Field
Parallel Plates - Slower Speed
- steeper curve
- shorter path
Fleming’s Left Hand Rule
Thumb - F - Force
First Finger - B - Field
Second Finger - I - Current
Right Hand Rule
Fingers - Current
Thumb - points in direction of North Pole
Magnetic Flux Density
Amount of magnetic flux through a unit area taken perpendicular to the direction of the magnetic field
Tesla
1 tesla is 1 Weber per metre squared
What is the equation for the force acting on a current in a magnetic field?
F = BILsinθ
l = current L = length of wire in the field θ = angle between the field lines and the wire
What is the equation for a charge moving perpendicular to a uniform magnetic field?
F = BQv
Mass Spectrometer
Particle accelerated through a potential difference
Particle passes through a velocity selector
Particle passes into the detector
Mass Spectrometer - Potential Difference
Particle is accelerated through a potential difference
W = VQ = 1/2 mv²
Mass Spectrometer - Velocity Selector
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.
Mass Spectrometer - Detector
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
Magnetic Flux
magnetic flux density x cross sectional area perpendicular to the field
Weber
One weber is the magnetic flux when a magnetic field of flux density 1T passes at right angles through an area of 1m²
What is the equation for magnetic flux?
ϕ = BA cosθ
Magnetic Flux Linkage
Magnetic flux density x number of turns on the coil x area of coil perpendicular to field
Faraday’s Law
E.M.F. Induced is proportional to the rate of change of flux linkage
E.M.F. = ∆Nϕ/ t
Lenz’s Law
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
AC Generator
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
Transformer - Turns Ratio
r = Vs/Vp = Ns/Np = Ip/Is
Describe the Function of a Simple Transformer
- 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
Why can’t DC be used for a transformer?
- 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
Transformers - Efficieny
Not 100% but can be up to 99% efficient
Eddy Currents
- 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
Inefficiencies in Transformers
Resistance of wires
Leakage inductance
Electric Field Strength
Force per unit positive charge
