background knowledge magnetic field

Background knowledge Magnetic field

Magnetic field can be produced by a magnet or a current-carrying conductor.

Magnetic field lines

Magnetic field lines are used to show the strength and direction of a magnetic field.

N SN S

Note:1. Field lines run from the N-pole round to the S-pole.2. When field lines are closely-spaced, field is strong and vice-versa.

N NN N

Neutral point

x

Example 1 (a)

In each of the following cases, draw the magnetic field lines between the magnets and mark the positions of the neutral points (if any).

S N

Example 1 (b)


N S

Example 1 (c)


S S

Example 1 (d)


Earth’s magnetic field

The Earth has a weak magnetic

field. It is so weak that it does

not affect much the field patterns

of permanent magnet.

It is rather like the field around a

huge bar magnet.

The south pole of the ‘earth

magnet’ is actually in the north.

Earth’s magnetic field

It is convenient to resolve the

earth’s field strength B into

horizontal and vertical component,

BH and BV respectively.

B

BH

BV

N

ground

We have BH = Bcos and BV = Bsin .Compass needles, whose motion is confined to a horizontal plane, are affected by BH only.

Fleming’s left-hand rule

F: force B: magnetic field I: current

F

B

I

Magnetic forcecurrent

Factors affecting magnetic force

B: magnetic field I: current l: length of conductor in B-field

F

B

I

Magnetic forcecurrent

Current balance

magnetic force F acting upwards

weight of the rider is equal to the magnetic force when the frame PQRS is balanced.

magnetic force and current

Record the

currents required to

balance different

number of identical

riders on PQ.

Results: magnetic

force (number of

riders) current

i.e. F I

magnetic force and length

Keep the current through PQ constant. Record the number of riders required to balance the metal fram

e when one, two and three pairs of equally strong magnadur magnets are used as shown.

Results: The magnetic force (number of riders) length of conductor (number of pairs of magnadur magnets)i.e. F l

magnetic force and magnetic field

Results: The magnetic force (number of riders) magnetic field (coil current) i.e. F B

Replace the magnadur magnets by a coil of many turns (for example 1100 turns) as shown below.

Hence, the magnetic field is provided from the coil and the magnetic field can be increased by increasing the coil current.

Keep the current through PQ constant. Record the coil currents required to bala

nce different number of identical riders on PQ.

Conclusion

The magnetic force is increased if the current is increased, (F I)

the magnetic field is increased, (F B) there is a greater length of wire inside the

magnetic field. (F l)

Precautions of using current balance

Make sure the direction of the magnetic field is perpendicular to the current-carrying arm.

Minimize the effect of the earth’s magnetic field by aligning the current-carrying arm along the N-S direction.

The set-up should be far from any current-carrying conductors so as to avoid the effect of stray magnetic fields.

To avoid overheating, the current should be switched off as soon as measurements have been taken.

Shield the set-up from the disturbance of wind.

Magnetic flux density B

Definition: Electric field E: force per unit charge

(E = F / Q) Gravitational field g: force per unit mass.

(g = F / m) Magnetic flux density B (magnetic field): force

acting per unit current length. (B = F / Il )

In words: The magnetic flux density B is equal to the force acting on a conductor of unit length and carrying a unit current at right angles to the field.

Unit of B: Tesla (T) or N A-1 m-1

Typical Values of the magnetic flux density

Source Magnetic field / T

Smallest value in a magnetically shielded room

10-14

Interstellar space 10-10

Earth's magnetic field 5 x 10-5

Small bar magnet 0.01

Big electromagnet 1.5

Strong lab magnet 10

Surface of a nucleus 106

Magnetic force on a current-carrying conductor

From definition: B = F/Il F = BIl

If the conductor and field are not at right angles, but make an angle with one another, the expression becomes

F = BIl sin .

Note that:1. when = 90o (conductor field), ⊥ F = BIl.2. when = 0o (conductor // field), F = 0.

B

Il

I cos

I sin

Magnetic force on moving charge in a magnetic field

Current = Q / t

conductor containing n particles each having charge q

sectionathroughpasstochargesallforrequiredtime

conductortheinchargetotal

l

nqv

vl

nqI

Magnetic force on moving charge in a magnetic field

Magnetic force acting on each moving charge

If the conductor makes an angle with the magnetic field,

F = Bqv sin

conductor containing n particles each having charge q

n

conductortheonactingforcetotal

n

ll

nqvB

n

BIlF

sinsin

Motion of a charged particle in a magnetic field

1 Moving parallel to the field

When a charged particle moves parallel to the magnetic field, i.e. = 0o, there is no force acting on it. Its velocity remains unchanged and its path is a

straight line.

vMagnetic field B+q

2 Moving perpendicular to the field

Radius of circular path

Circular path of motion of +q

uniform magnetic field B


+q+q


Period of circular motion




+q+q


Notes: The magnetic force is always perpendicular to the motion of the

charged particle; no work is done by the magnetic field. Kinetic energy of the charged particle remains constant.




+q+q

Example 2 The path of an electron in a uniform magnetic field of flux density

0.01 T in a vacuum is a circle of radius 0.05 m. Given that the charge and mass of an electron are -1.6 x 10-19 C and 9.1 x 10-31 kg respectively. Find (a) the speed and(b) the period of its orbit.

Solution:

Current balance

To measure steady magnetic field by using principle of moment.

Clockwise moment = anti-clockwise moment

mgd2 = BIld1

rider

Magnetic forceBIl

mechanical forcemg

d1 d2

Example 3 Current balanceA rider of mass 0.084 g is required to balance the frame when an arm PQ, of length 25 cm and carrying a current of 1.2 A, is inside and in series with a flat, wide solenoid as shown below.Find the magnetic flux density inside the solenoid.

Search coil and CRO

A search coil is only used in measuring a varying magnetic field.

A typical search coil consists of 5000 turns of wire and an external diameter d ≤ 1.5 cm so that it samples the field over a small area.

The search coil is placed with its plane perpendicular to a varying magnetic field.

Search coil and CRO

Due to the change of magnetic flux, an e.m.f., which is induced in the search coil, can be measured by a CRO.

The induced e.m.f. E is proportional to the flux density B and the frequency f of the varying magnetic field. i.e. E Bf

The earth’s magnetic field can be ignored because it is a steady field.

The CRO is a perfect voltmeter as its resistance is very high. It can measure both d.c. and a.c. voltages and show how they vary with time.

Cathode-ray oscilloscope (CRO)

simulation

Measuring d.c. voltages

To measure a d.c. voltage, the time base is usually switched off. Thus, it is the light spot which is deflected.

Or the time base may be switched on to any high value so that the horizontal trace is deflected. From the deflection on the screen and the gain control setting, the d.c. voltage is then calculated.

Y-gain control:2 V cm-1

Vd.c. = amount of deflection (cm) x Y-gain control setting (V cm-1) = 3 x 2 = 6 V

Measuring a.c. voltages To measure an a.c. voltage, the time base is usually switch

ed off. The waveform displayed becomes a vertical trace so that the amplitude can be easily read on the screen. The maximum voltage or peak voltage is then calculated from the gain control setting. Note that the peak voltage refers to half the length of the vertical trace.

Y-gain control:5 V cm-1

Time base setting:10 ms cm-1

Peak voltage =

Period =

Frequency =

Example 4CRO waveformThe figure below shows a waveform on a screen.(a) If the controls on the CRO are set at 0.5 V cm-1 and 2 ms cm-1,

(i) the peak voltage, and (ii) the frequency of the input signal.

Solution:

1 cm

1 cm

Example 4CRO waveformThe figure below shows a waveform on a screen.(b) If the gain control is changed to 1 V cm-1, sketch the trace on the figure.Solution:

1 cm

1 cm

Example 4CRO waveformThe figure below shows a waveform on a screen. (c) If the time base control is changed to 5 ms cm-1, sketch the trace on the figure.Solution:

1 cm

1 cm

Time base

When Vx increases linearly with time from A to B, the spot of light on the screen moves at a constant speed from the left to right of the screen.

Then the spot of light flies back to the left quickly when Vx suddenly drops from B to C.

The saw-tooth voltage Vx causes no vertical movement of the spot of light..

Example 5 displaying a.c. waveforms

Measuring of phase relationships The phase difference of two p.ds (Vx and

Vy) can be observed on phase difference a double-beam CRO.

phase difference

If a double-beam CRO is not available, the phase difference can be found by applying the two p.ds of the same frequency and amplitude to the X- and Y-plates (time base off) simultaneously.

The phase difference can be determined from the trace on the screen of the CRO as follows.

In general, the trace is an ellipse except when the f is 0o, 90o, 180o, 270o or 360o.

Comparing of frequencies When two p.ds of different frequencies fx and fy are applie

d to the X- and Y-plates (time base off), more complex figures are obtained, know as Lissajous’ figures.

simulation

In any particular case, the frequency ratio can be found by

lineverticalltouchingloopsofno.

linehorizontaltouchingloopsofno.

x

y

f

f

Magnetic field around a long straight wire

1. The field lines are circles around the wire.

2. The magnetic field is the strongest close to the wire.

3. Increasing the current makes the magnetic field stronger.4. Reversing the current also

reverses the direction of field lines, but the field pattern

remains unchanged.

Right hand grip rule

If the right hand grips the wire so that the thumb points the same way as the current, the fingers curls the same way as the field lines.

Experiment to show that B ∝ I and B ∝ 1 / r

B ∝ I and B 1 /∝ r B ∝ I/r B = 0I/(2r) where I is the current and is the permeability of free

space (0 = 4 x 10-7 T m A-1)

B = 0I/(2r) (0 = 4 x 10-7 T m A-1)

Example 7 Two long wires X and Y each carries a current

of 20 A in the directions as shown in the figure. If the distance between the wires is 10 mm, find the magnitude and direction of the magnetic flux density at

(a) P P Q

X Y

I = 20 A I = 20 A

5 mm 5 mm 5 mm

B = 0I/(2r) (0 = 4 x 10-7 T m A-1)

Example 7 Two long wires X and Y each carries a current

of 20 A in the directions as shown in the figure. If the distance between the wires is 10 mm, find the magnitude and direction of the magnetic flux density at

(b) Q. P Q

X Y

I = 20 A I = 20 A

5 mm 5 mm 5 mm

Magnetic field around a flat coil

At the centre of the coil The field lines are straight an

d at right angles to the plane of the coil

Outside the coil, The field lines run in loops.

Experimental set-up

By using the experimental-setup shown, it is found that the magnetic field is 1. directly proportional

to the current and the number o

f turns, and 2. inversely proportional to the radius of the

coil.

Note: The magnetic field is greatest at the centre.

r

NIB

20

Magnetic field due to a long solenoid

From the field pattern of the solenoid, it can be found that1. inside the solenoid, the field lines are straight and

evenly-spaced. This indicates that the field is of uniform strength.

2. outside the solenoid, the pattern is similar to that around a bar magnet, with one end of the solenoid behaving like a N-pole and the other end like a S-

pole.

I I

NS

Right hand grip rule

If the right hand grips the solenoid so that the fingers curls the same way as the current, the thumbs points to the north pole of the solenoid.


The magnetic field of the solenoid can be increased by

1. increasing the current,

2. increasing the number of turns on the coil.

I I

NS


The magnetic field of the solenoid can be increased by 1. increasing the current, 2. increasing the number of turns on the coil.

In vacuum, the magnitude of the magnetic flux density B at a point O on the axis near the centre of the solenoid of length l, having N turns and carrying a current I is given by

B = 0NI/l or B = 0nI where n is the number of turns per unit length (n = N/l)

I I

NS

Experiment to show that B ∝ N, B ∝ I and B 1 /∝ l


Note: The magnetic field within a solenoid is independent of the shape and the cross section area of the solenoid.

I I

NS


At the ends of solenoid The magnetic field at the ends of the solenoid is weaker. It is

half that in the central region within the solenoid.

222

1' 00 nI

l

NIBB

r

IB

2

0r

NIB

20

nIl

NIB 0

0

l

Nn

2200 nI

l

NIB

N = number of turnsl = length of solenoid

Current-carrying conductor

Position of magnetic field

Magnetic flux density (B)

Symbol

1. Long straight wireAround the wire

r = perpendicular distance from wire

2. Circular coil At the centreN = number of turnsr = radius of coil

3. Solenoid

Inside

At the ends

Force between currents

Magnetic field due to current through P

Magnetic force acting on Q

QP

B1

F2

r

IB

2

101

r

lIIlIBF

2210

212

Applying Fleming’s left hand rule, force acting on wire Q is towards wire P.

B2

F1PQ

r

lIIF

2210

1

Similarly, the force acting on wire P is towards wire Q. Hence, the two wires attract each other.

By Newton’s third law, the magnetic force acting on P = magnetic force acting on Q.

∴

Summary: Unlike current repel, like current attract

Parallel wires with current flowing in the same direction, attract each other.

Parallel wires with current flowing in the opposite direction, repel each other.


The force per unit length on each conductor

When the current I1 = I2 = 1 A, and the separation

between the wires r = 1 m,

the force per unit length on the conductor

r

lIIFF

2

21021

r

II

l

F

l

F

2

21021

77

10212

11104

N m-1


Definition of the ampere The ampere is constant current which, flowing in two infinit

ely long, straight, parallel conductors of negligible cross-section and placed in a vacuum 1 metre apart, produces between them a force equal to 2 x 10-7 Newton per metre of their length.

r

lIIlIBF

2210

212

Example 8

The figure below shows two horizontal wires, P and Q, 0.2 m apart, carrying currents of 1.5 A and 3 A respectively.

(a) Calculate the force per metre on wire Q. (b) State the direction of the force. (c) State the direction and magnitude of the force per metre

on wire P due to the current in wire Q

.

Moment and couple

Couple ─ consists of 2 equal and opposite parallel forces whose lines of action do not coincide ( 重疊 ).

torque of couple = F x d/2 + F x d/2 = Fd

F

Fd/2 d/2

F

F

dd

Torque on a rectangular current-carrying coil in a uniform B-field

Torque = F(b sin )= NBIlb sin = NBAI sin

F’

F’

Maximum and minimum torque on a coil

The maximum torque is NBAI when the plane of coil is parallel to the field ( = 90o).

The torque on the coil is zero when the plane of coil is perpendicular to the field ( = 0o).

Example 9

Example 9 A square coil has sides of length 5 cm. The coil consis

t of 20 turns of insulated wire carrying a current of 0.2 A. The plane of the coil is at an angle 40o to a uniform magnetic field of flux density of 25 mT. Calculate the torque acting on the coil.

Solution:

Moving-coil galvanometer

The moving-coil meter contains a coil wound on an aluminium former around a soft-iron cylinder.

The coil is pivoted on bearings between the poles of a cylindrical magnet.

Current flows through the coil via a pair of spiral springs called hair springs.

Theory

When a current is passed through a coil in a magnetic field, the coil experiences a torque. The coil rotates, moving the pointer across the scale.

The normal of plane of the coil is always perpendicular to the magnetic field, the torque on the coil is given by

T = NBAI sin 90o = NBAI

Theory

The movement of the coil is opposed by the hair springs.

The restoring torque () exerted by the hair springs to oppose the rotation is given by

= k where k is the torsion

constant of the hairsprings

Linear scale

The coil comes to rest when the magnetic turning effect (torque) on the coil is balanced by the turning effect (restoring torque) from the hair springs.

BANI = k Hence, I ∝ ; the galvanometer scale is linear

Sensitivity

The current sensitivity of a galvanometer is defined as the deflection per unit current

i.e. current sensitivity = /I = BAN/k. The voltage sensitivity of a galvanometer is defined as the deflect

ion per unit voltage

i.e. voltage sensitivity = /V = /(IR) = BAN/(kR)

where R is the resistance of the coil. The sensitivity can be increased, i.e. the coil deflects more for a c

ertain current or voltage, by

1. using a stronger magnet (larger B),

2. using weaker hair springs (smaller k).

3. using a coil with larger area (larger A), and

4. increasing the number of turns of the coil (larger N).

background knowledge magnetic field

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