MAGNETIC FIELDS

?
  • Created by: CPev3
  • Created on: 06-04-21 17:04

Magnetic field

Field surrounding a permanent magnet/ current-carrying conductor 

...in which magnetic objects experience a force

1 of 27

Magnetic field lines

Lines of force used to map the magnetic field pattern

...around a permanent magnet/ current-carrying conductor

2 of 27

Properties of magnetic field lines

  • Arrow = direction in which a free north pole would move
    • Points from north to south

.

  • Equally spaced and parallel = uniform magnetic field
    • Strength of the magnetic field does not vary

.

  • Closer together = stronger magnetic field
    • For a bar magnet, the magnetic field is strongest at its north and south poles
3 of 27

How to detect the presence of a magnetic field

The needle of a small plotting compass

......will deflect in the presence of a magnetic field

.

A magnetic field around a bar magnet

......induces magnetism in iron filings

............which line up in the field

4 of 27

Electromagnetism

When a wire carries a current, a magnetic field is produced around the wire

The magnetic field is created by the electrons moving within the wire

Any charged particle that moves creates a magnetic field in the space around it

.

A magnetic field is produced around a bar magnet

The magnetic field is created by the electrons moving around the iron nuclei

5 of 27

Current-carrying wires

Magnetic field lines

  • Concentric circles
  • Centred on the wire
  • Perpendicular to the wire

.

Current out of the plane of the paper

  • Anticlockwise
  • Represented by a dot

.

Current into the plane of the paper

  • Clockwise
  • Represented by a cross
6 of 27

Right-hand grip rule

Thumb = direction of the conventional current

Curled fingers = direction of the magnetic field

7 of 27

When forces are experienced

When a wire carries a current, a magnetic field is produced around the wire

The wire can be placed in an external magnetic field

The two fields interact just like the fields of two permanent magnets

The two magnets experience equal and opposite forces

.

When a current-carrying wire is placed between the poles of a magnet

  • Each pole experiences a force 1/2 F
  • The wire experiences a force F in the opposite direction
8 of 27

Fleming's left-hand rule

First finger = direction of the external magnetic field

Second finger = direction of the conventional current

Thumb = direction of motion (force) of the wire

9 of 27

Magnitude of force

The magnitude of the force experienced by a current-carrying wire in an external magnetic field

......is a maximum when the wire is perpendicular to the magnetic field

............and zero when the wire is parallel to the magnetic field

10 of 27

Force equation

F = BIL sinθ

  • F = force
  • B = magnetic flux density (strength of the magnetic field)
  • I = current
  • L = length of the wire in the magnetic field
  • θ = angle between the magnetic field and current

.

When the wire is perpendicular to the magnetic field

  • θ = 90o and sin90 = 1
  • F = BIL
11 of 27

Magnetic flux density

The strength of a magnetic field

.

The magnetic flux density is 1 T

...when a wire carrying a current of 1 A placed perpendicular to the magnetic field

......experiences a force of 1 N per metre of its length

.

1 T = 1 Nm-1A-1

12 of 27

Determining magnetic flux density

Two magnets placed on a top-pan balance

Uniform magnetic field between the magnets

Stiff copper wire held perpendicular to the field

Length of the wire measured with a ruler

Wire connected in series with an ammeter and variable power supply

No current: balance zeroed

Current: wire experiences a vertical force

Magnets experience an equal and opposite force

F = mg

B = F / IL

13 of 27

Electron deflection tube

Uniform magnetic field

Beam of electrons moving from left to right

Experiences a downward force when it enters the field

Travels in a circular path

Force perpendicular to velocity

Speed unchanged

Moves in a straight line when it exits the field

14 of 27

Force equation

When a charged particle moves perpendicular to a uniform magnetic field

.

Fconductor = BIL

L = vt

F = BIvt

.

I = NQ / t

F = BNQvt / t = BNQv

.

Fcharged particle = NBQv / N = BVQ

.

Felectron = Bev 

15 of 27

Radius equation

F (centripetal) = mv2 / r

F (magnetic) = BQv

mv2 / r .= BQv

r = mv / BQ

16 of 27

Velocity selector

Uses electric and magnetic fields to select charged particles of a specific velocity

.

Two parallel horizontal plates connected to a power supply

Uniform electric field between them

Perpendicular to a uniform magnetic field

Charged particles enter narrow slit Y

Fields deflect them in opposite directions

Fields cancel for particles with a specific velocity

These particles move in a straight line and exit narrow slit Z

17 of 27

Velocity selector equation

F (electric force) = EQ

F (magnetic force) = BQv

EQ = BQv

v = E / B

= (4 x 105) / 0.1

= 4 x 106 ms-1

18 of 27

Inducing an electromotive force

  • Stationary = no reading on voltmeter
  • Magnet pushed towards coil = electromotive force induced across ends of coil
  • Magnet pulled away from coil = reverse electromotive force induced across ends of coil
  • Repeated pushing and pulling = alternating current induced
  • Faster pushing and pulling = larger electromotive force induced
19 of 27

Magnetic flux equation

Φ = BAcosθ

  • Φ = magnetic flux
  • B = magnetic flux density
  • A = cross-sectional area
  • θ = angle between the magnetic field and normal
20 of 27

Magnetic flux linkage equation

Magnetic flux linkage = NΦ

  • N = number of turns in the coil
  • Φ = magnetic flux

.

Change in magnetic flux linkage (B, A or θ) = induced electromotive force

21 of 27

Transformer

  • Laminated soft iron core
  • Primary (input) coil
  • Secondary (output) coil
  • Alternating current supplied to the primary coil
    • Produces a changing magnetic flux in the core
    • Core ensures the magnetic flux produced by the primary coil links the secondary coil
    • No magnetic flux is lost
    • Produces a changing electromotive force in the secondary coil (Faraday's law)
22 of 27

Turn-ratio equation

ns / np = Vs / Vp

  • ns = number of turns on the secondary coil
  • np = number of turns on the primary coil
  • Vs = output voltage
  • Vp = input voltage
23 of 27

Types of transformer

Step-up

  • ns > np
  • Vs > Vp

.

Step-down

  • ns < np
  • Vs < Vp
24 of 27

Investigating transformers

Apparatus

  • Iron C-core
  • Thin insulated copper wires for the primary and secondary coils
  • Signal generator in series with the primary coil
  • Multimeter set to alternating voltage in parallel around each coil

.

Method

  • Change the number of turns on one or both coils
  • See what happens to Vs for a fixed value of Vp and vice versa
25 of 27

100 % efficient transformers

Pp  = Ps

IpVp = IsVs

Ip / Is = Vs / Vp

.

Step-up transformer: ↑ V = ↓ I = constant P

Step-down transformer: ↓ V = ↑ I = constant P

26 of 27

How to make a transformer efficient

Low-resistance wires

  • Minimise power losses due to the heating effect of the current

.

Laminated core

  • Layers of iron separated by an insulator
  • Minimises currents induced in the core
  • Minimises power losses due to the heating effect of the current

.

Soft iron core

  • Easy to magnetise/ demagnetise
27 of 27

Comments

No comments have yet been made

Similar Physics resources:

See all Physics resources »See all Magnetic fields resources »