Physics 12 Lesson Notes: Natural Magnetism and Electromagnetism

MagnetsWhen a bar magnet is dipped into iron filings, the filings are attracted to it, accumulating

most noticeably around regions at each end of the magnet�the poles. When the bar magnet isallowed to rotate freely the pole that tends to seek the northerly direction is called the north-seeking pole, or simply, the N-pole. The other is called the south-seeking pole, or S-pole.

By placing two bar magnets first with similar poles together, then with opposite polestogether, you can demonstrate the law of magnetic poles (Figure 1):

Magnetic FieldsSince an iron filing experiences a force when placed near a magnet, then, by definition,

a magnet is surrounded by a magnetic force field. This field is often detected by its effect on asmall test compass (magnetized needle). It is visually depicted by drawing magnetic field linesthat show the direction in which the N-pole of the test compass points at all locations in the field.Experimentally, the lines in a magnetic field can easily be traced by sprinkling iron filings on asheet of paper placed in the field. The filings behave like many tiny compasses and line up in thedirection of the field at all points. They produce a �picture� of the magnetic field, as shown inFigure 2.

Since iron filings have no marked north or S-poles, they reveal only the pattern of themagnetic field lines, not their direction (Figure 3). The relative strength of the magnetic field isindicated by the spacing of adjacent field lines: where lines are close together, the magnetic fieldis strong.

(a) Similar poles face eachother.

(b) Opposite poles face eachother.

The magnetic field at any point is a vector quantity, represented by the symbol B. Themagnitude B is given by the magnitude of the torque (or turning action) on a small test compassnot aligned with the direction of the field. We will make a more precise definition of B later inthis chapter, when we examine electromagnetism.

The characteristics of magnetic field lines are summarized below.

1. The spacing of the lines indicates the relative strength of the force. The closertogether the lines are, the greater the force.

2. Outside a magnet, the lines are concentrated at the poles. They are closestwithin the magnet itself.

3. By convention, the lines proceed from S to N inside a magnet and from N toS outside a magnet, forming closed loops. (A plotting compass indicates these directions.)

4. The lines do not cross one another.

Note that the magnetic field around a barmagnet is three-dimensional in the diagram shown;it does not exist just in the horizontal plane.

Earth’s Magnetic FieldA pivoted magnet will rotate and point north�south because of its interaction with the

magnetic field of Earth. As early as the 16th century, Sir William Gilbert, the distinguishedEnglish physicist, had devised a model to describe Earth�s magnetism. He determined thatEarth�s magnetic field resembled the field of a large bar magnet, inclined at a slight angle toEarth�s axis, with its S-pole in the northern hemisphere. Figure 4(a) shows this field and the barmagnet that was thought, in Gilbert�s time, to be responsible for it.

(a) The magnetic field of Earthclosely resembles the field of a largebar magnet.

(b) Lines of magnetic declination inCanada

A compass points toward Earth�s magnetic S-pole, rather than toward its geographic northpole (the north end of Earth�s axis of rotation). The angle, or magnetic declination, betweenmagnetic north and geographic north varies from position to position on the surface of Earth(Figure 4(b)). In navigating by compass, the angle of declination for a particular location mustbe known so that true north can be determined.

In addition, Earth�s magnetic field is three-dimensional, with both a horizontal and avertical component. A magnetic compass on a horizontal surface reveals only the horizontalcomponent. The angle between Earth�s magnetic field, at any point, and the horizontal is calledthe magnetic inclination, or �dip,� and is measured with a magnetic dipping needle (Figure 5).

A dipping needle is a compass pivotedat its centre of gravity and free to rotate in a vertical plane. Whenaligned with a horizontal compass pointing north, it points in the direction ofEarth�s magnetic field. The angle of inclination is then read directly from theattached protractor.

Inclination and declination charts must be revised from time to time because Earth�smagnetic field is slowly changing. It is believed that these changes result from the rotation of themagnetic field about Earth�s axis; one complete rotation takes about 1000 years (Figure 6).

The Domain Theory of MagnetismAlthough not normally magnetized, some ferromagnetic materials, such as iron,

nickel, cobalt, and gadolinium, may become magnetized under certain circumstances. How theyare able to acquire magnetic properties may be explained by the domain theory of magnetism.

Ferromagnetic substances are composed of a large number of tiny regions calledmagnetic domains. Each domain behaves like a tiny bar magnet, with its own N- and S-poles. Whena specimen of the material is unmagnetized, these millions of domains are oriented at random, withtheir magnetic effects canceling each other out, as in Figure 7.

However, if a piece of ferromagnetic material is placed in a sufficiently strongmagnetic field, some domains rotate to align with the external field, while others, already aligned,tend to increase in size at the expense of neighboring nonaligned domains (Figure 8). The netresult is a preferred orientation of the domains (in the same direction as the external field),causing the material to behave like a magnet. When the external field is removed, this orientationwill either remain for a long time or disappear almost immediately, depending on the material.When magnets are made in this way, they are known as induced magnets.

Figure 7The atomic dipoles are lined up ineach domain. The domains point inrandom directions. The magneticmaterial is unmagnetized.

Figure 8The atomic dipoles (not the domains) turn so that alldomains point in the direction of the magnetizing field.The magnetic material is fully magnetized.

The domain model provides a simple explanation for many properties of induced magnets:

1. A needle is magnetized by rubbing it in one direction with a strong permanent magnet. Thisaligns the domains with the field of the permanent magnet.

2. When a bar magnet is broken in two, two smaller magnets result, each with its own N- and S-poles. It is impossible to produce an isolated N- or S-pole by breaking a bar magnet.

3. Induced magnets made of �soft� iron demagnetize as soon as the external field is removed.Examples include temporary magnets such as lifting electromagnets. In contrast, hard steel oralloys remain magnetized indefinitely. These include permanent magnets such as magneticdoor catches. Impurities in the alloys seem to �lock� the aligned domains in place andprevent them from relaxing to their random orientation.

4. Heating or dropping a magnet can cause it to lose its magnetization, jostling the domainssufficiently to allow them to move and resume their random orientation. Each ferromagneticmaterial has a critical temperature above which it becomes demagnetized and remainsdemagnetized even upon cooling.

5. A strong external magnetic field can reverse the magnetism in a bar magnet, causing theformer south-seeking pole to become north-seeking. This occurs when the domains reversetheir direction of orientation by 180° due to the influence of the strong external field in theopposite direction.

6. Ships� hulls, columns and beams in buildings, and many other steel structures are often foundto be magnetized by the combined effects of Earth�s magnetic field and the vibrationsimposed during construction. The effect is similar to stroking a needle with a strong magnet,in that the domains within the metals are caused to line up with Earth�s magnetic field.Vibrations during construction aid in the realignment of the domains.

Example 1) Two iron nails are held to a magnet, as shown in Figure. Predict what will happenwhen the nails are released. If possible, verify your prediction experimentally.

Example 2) In the diagrams below, each circle represents a compass. Show the direction of theneedle in each compass.

Example 3) In the diagrams below, draw the magnetic field lines between the ends of three barmagnets if(a) all the S-poles are close together(b) one N-pole and two S-poles are close together

Small pieces of iron rubbed in one direction with lodestone become magnetized. Evenbringing a piece of iron near a magnet causes the iron to be magnetized. Nickel and cobalt, andany alloy containing nickel, cobalt, or iron, behave in the same way. These substances are calledferromagnetic, and you can induce them to become magnetized by placing them in a magneticfield.

Oersted’s Discovery

For centuries, people believed that electricity and magnetism were somehow related, butno one could prove a connecting link between them. Then, in 1819, the Danish physicist HansChristian Oersted (1777�1851) discovered the connection by accident while lecturing on electriccircuits at the University of Copenhagen.Oersted noticed that a compass needle placed just belowa wire carrying a current would take up a position nearly perpendicular to the wire while thecurrent was flowing (Figure 1). When the direction of the current was reversed, the compassneedle again set itself at right angles to the wire, but with its ends reversed. The effect lasted onlywhile the current flowed. Much to his own surprise, Oersted had discovered the basic principleof electromagnetism.

Magnetic Field of a Straight ConductorWhen an electric current flows through a long, straight conductor, the resulting

magnetic field consists of field lines that are concentric circles, centred on the conductor(Figure 9). You can remember the direction of these field lines (as indicated by the N-pole of asmall test compass) if you use the right-hand rule for a straight conductor:

(a) Iron filings reveal the circular patternof the magnetic field around a conductorwith a current.

(b) If the right thumb points in thedirection of the current, thenthe fingers curl around the wirein the direction of the magneticfield lines.

Models of the magnetic field of a straight conductor(a) Imagine the X as being the tail of an arrow moving away from you.(b) Imagine the dot as being the tip of an arrow facing you.

Understanding Concept.

1) Refer to diagram on the right and fill in the needles by showingwhich direction are they pointing if current is applied in thecircuit?

2) Figure shows three current-carrying conductors with theirmagnetic fields. Indicate the direction of electric current ineach wire.

3) Figure shows three conductors with the direction of the electric current. Draw magnetic fieldlines around each, indicating polarities where applicable.

4) Choose the diagram from below that best illustrates the strength of the magnetic fieldsurrounding a conductor. Explain your answer.

Determine the magnetic field around a current carrying straight wire

To determine the magnitude of the magnetic field strength, we use the formula

Where B is the magnetic field strength generated by the current.μo is a constant, called the permeability of free space,

μo = 4 π x 10�7AmT

I is the current through the wireR is the distance to the conductor.

Example 1)What is the magnetic field strength 15 cm from a straight conductor with a current of60 A flowing through it?

Example 2) What current is flowing through a straight wire if the magnetic field strength 35 cmfrom the wire is 6.4 x 10� 6 T?

Example 3) What is the magnetic field strength at a point midway between two long parallelwires, 1.0 m apart, carrying currents of 10 A and 20 A respectively, if the currents areflowing in the same direction?

Example 4) What is the magnetic field strength at a point midway between two long parallelwires, 1.0 m apart, carrying currents of 10 A and 20 A respectively, if the currents areflowing in the opposite direction?

Understanding the concepts.

1) What is the magnetic field strength 20 cm from a ling straight conductor with a current of 60A flowing through it? 6.0 x 10� 5 T

2) What current is flowing through a straight wire if the magnetic field strength 10 cm from thewire is 2.4 x 10� 5 T? 12 A

3) At what distance from a straight conductor, with a current of 200 A flowing through it, is themagnetic field intensity 8.0 x 10� 4 T? 5.0 x 10� 2 m.

4) What is the magnetic field strength at a point midway between two long parallel wires, 1.0 mapart, carrying currents of 10 A and 20 A respectively, if the currents are

a) in opposite directions? 1.2 x 10� 5 Tb) in the same direction? 4.0 x 10� 6T

5) A long, solid, copper rod has a circular cross-section of diameter 10 cm. It carries a current of1000A, uniformly distributed across its area. Calculate the magnetic field strength at thesefour positions: (Tricky!!!)

a) at the centre of the rod 0Tb) 2.5 cm from the centre 2.0 x 10� 3Tc) 5.0 cm from the centre 4.0 x 10� 3Td) 7.5 cm from the centre 2.7 x 10� 3T

Magnetic Field of a Current Loop

When a straight wire is formed into a circular loop, its magnetic field will appear asshown in Figure 10. Note that the field lines inside the loop are closer together, indicating astronger magnetic field than on the outside of the loop.

Magnetic Field of a Coil or SolenoidA solenoid is a long conductor wound into a coil of many loops. The magnetic field

of a solenoid (Figure 11) is the sum of the magnetic fields of all of its loops. The field inside thecoil can consequently be very strong. If the coil is tightly wound, the field lines are nearlystraight and very close together (Figure 12).

Figure 11(a) A solenoid

(b) Iron filings reveal the fieldlinesin and around a solenoid.

Figure 12

(a) When the solenoid is loosely wound, field lines within the coil are curved.

(b) The field becomes stronger and straighter inside the coil when the coil is woundtighter. The right-hand rule for solenoids (a corollary of the right-hand rule for astraight conductor) gives the direction of the field inside the coil.

A solenoid has a magnetic field very similar to the field of a bar magnet, with theconvenient additional feature that the field can be switched off and on. To remember thedirection of the magnetic field of a solenoid, we apply a special right-hand rule for a solenoid:

Note that the right-hand rule for a solenoid is consistent with the right-hand rulefor a straight conductor if we point our thumb along, or tangent to, the curved wireof the coil.

Determine the magnetic field at the centre of a current carrying circular loops

If the current carrying conductor is bent into a circular loop, the magnetic field at thecentre of the loop is found using the formula:

,Example 1) A circular coil has 10 loops of wire with a diameter of 12 cm. If the current flowing

through these loops is 5.0 A, what is the magnetic field strength at the centre of this coil?

Example 2) What is the magnitude of the magnetic field in the core of a solenoid 5.0 cm long,with 300 turns and a current of 8.0 A?

Example 3) An air core solenoid is 25 cm long and has 1000 loops. If it has a current of 9.0 A,what is the magnetic field in this solenoid?

Understanding Concepts.

1) A circular coil has a diameter of 9.0 cm and 12 loops. If the current flowing through the coil is15 A, what is the magnetic field strength at the centre of this coil? 2.5 x 10�3 T

2) A 25.0 cm solenoid has 1800 loops and a diameter of 3.0 cm. Calculate the magnitude of themagnetic field in the air core of the solenoid when a current of 1.25 A is flowing. 1.13 x 10�2 T

3) A circular coil has 9 loops and a current of 8.0 A flowing through it. If the magnetic field atthe centre of this coil is 1.1 x 10�3 T, what is its diameter? 0.082 m

4) A circular coil with 18 loops of wire and has a diameter of 12 cm. If the magnetic field at thecenter of this coil is 6.2 x 10�4 T, what is the current flowing through the coil? 3.3 A

5) An air core solenoid is 25 cm long and carries a current of 0.72 A. If the magnetic field in thecore is 2.1 x 10�3 T, how many turns does this solenoid have? 580

6) An air core solenoid is 30.0 cm long and has 775 turns. If the magnetic field in the core is0.100 T, what is the current flowing through this solenoid? 31 A

7) Two long fixed parallel wires are 7.2 cm apart and carry currents of 25 A and 15 A in theopposite direction. What is the magnitude of the magnetic field midway between the twowires? 2.2 x 10�4 T

8) A 14-gauge copper wire has a current of 12 A. How many turns would have to be wound on acoil 15 cm long to produce a magnetic field of strength 5.0 x 10�2 T? 5.0 x 102

9) Calculate the magnitude of the magnetic field strength in the core of a coil 10.0 cm long, with420 turns and a current of 6.0 A. 3.2 x 10�2 T

10) A coil 8.0 cm long, with 400 turns, produces a magnetic field of magnitude 1.4 x 10�2 T inits core. Calculate the current in the coil. 2.2 A

Physics 12 Electromagnetism.Lesson Notes: Magnetic Force on Moving Charge

Before we start this lesson, let us review the interactions of magnetic fields.

What happens in the case of a pair of parallel current carrying conductors?Case I) In the same direction.

Case II) In the opposite direction.

From the above diagrams, we have observed that the magnetic field�s set upplays a big role in the outcome of either force of repulsion or attraction. We recognize that thebigger the magnetic field exists between the wires, the bigger the magnetic force occurs.

To determine the magnitude of these attractive or repulsive forces between theconductors, we use the following formula:

WhereLF = force per unit length,

I1 and I2 = the currents in the conductors.

R = the distance between the conductors

Example 1) What is the magnetic force between two wires carrying currents of 5.0 A and 8.0 Ain opposite directions? If the wires are 15 cm apart and 45 cm long.

Example 2) Two long parallel wires carrying currents of 10.0 A and 15.0 A in the same direction.If the force per meter on each wire is 4.8 x 10�4 N/m, how far apart are the wires?

Behavior of a current carrying conductor in Magnetic Field.

In 1821, following Oersted�s discovery of electromagnetism, English physicist MichaelFaraday (1791�1867) set out to prove that, as a wire carrying electric current could cause amagnetized compass needle to move, so in reverse a magnet could cause a current-carrying wireto move. Suspending a piece of wire inside a magnetic field which came from a fixed magnet,Faraday connected the wire to a battery, and the wire began to deflect.

Faraday determined that the magnetic field of a permanent magnet can exert a force onthe charges in a current-carrying conductor. Figure 1 shows how the direction of this force isrelated to the magnetic field of the conductor and to the external magnetic field.

To the left of the conductor, the field lines point in the same direction and tend toreinforce one another, producing a strong magnetic field. To the right, the fields are opposed and,as a result, tend to cancel one another, producing a weaker field. This difference in field strengthresults in a force to the right on the conductor. If either the external field or the direction of theelectric current were reversed, the force would act in the opposite direction. A more detailedinvestigation would show that the actual magnitude of the force depends on the magnitude ofboth the current and the magnetic field. These effects are summarized in themotor principle.

Motor PrincipleA current-carrying conductor that cuts across external magnetic

field lines experiences a force perpendicular to both the magnetic fieldand the direction of electric current. The magnitude of this force dependson the magnitude of both the external field and the current, as well as theangle between the conductor and the magnetic field it cuts across.

The magnitude of the magnetic force on a current carrying conductor canbe calculated using:

Fm = B I L sin θ where Fm = magnetic forceB = magnetic field strengthI = current through the conductorL = length of conductor

sin θ = angle at which the conductor passes throughthe magnetic field

Example 1) A conductor 3.2 x 10�1m is placed in a magnetic field of 2.10 x 10�1 T. Assumingthe conductor is perpendicular to the magnetic field, and the magnetic force acting on theconductor is 4.00 x 10�2N, what is the current flowing through the conductor?

Example 2) Calculate the magnitude and the direction of the magnetic force on a conductorwhich is 5.5 cm long with a current of 4.0 A heading North. The conductor is placedperpendicular to the magnetic field with strength of 3.5 x 10�2 T from West to East.

Example 3) Calculate the magnitude and the direction of the magnetic force on a conductorwhich is 8.0 cm long with a current of 7.5 A. The conductor is placed at an angle 30° to themagnetic field with strength of 3.5 x 10�2 T from East to West.

Moving charges in Magnetic Fields.

It is not only current carrying conductors that are deflected by magneticfields. Moving charges (electrons, protons, and alpha particles) can also bedeflected by magnetic fields. These particles are deflected in the same way ascurrent carrying conductors. However, to determine the direction of the deflection,you use your left hand rule for negative charge (electrons) and right hand rule forall positive charges like you did with the current carrying wire in magnetic field.

The magnitude of the deflecting magnetic forces on moving charged particles canbe calculated using:

Fm = q v B sin θ where Fm = magnetic forceB = magnetic field strengthq = charge of the particlev = speed of the particle

sin θ = angle at which the particle passes through themagnetic field

Example 4) Calculate the magnitude of the magnetic force on an electron traveling at a speed of3.6 x 104 m/s perpendicular through a magnetic field 4.20 T. 2.42 x 10�14 N

Example 5) A proton traveling at an angle 35° to the vertical line at a speed of 2.1 x 10 5 m/sthrough a horizontal magnetic field experiences a magnetic force of 7.8 x 10�14 N.What is the magnitude of the magnetic field? 2.83 T

Figure 5 shows a positively charged particle in a magnetic field perpendicular to its velocity. (Ifthe particle were negative, its trajectory would curve the other way, in a clockwise circle.) If themagnetic force is the sole force acting on the particle, it is equal to the net force on the particleand is always perpendicular to its velocity. This is the condition for uniform circular motion; infact, if the field is strong enough and the particle doesn�t lose any energy, it will move in acomplete circle as shown.

Figure 5(a) A positive charge moving atconstant speed through a uniformmagnetic field follows a curved path.

(b) Ideally, a charged particle willmove in a circle because themagnetic force is perpendicularto the velocity at all times.

We represent these magnetic fields in two-dimensional diagrams by drawing Xs for field linesdirected into and perpendicular to the page and dots for field lines pointing out of andperpendicular to the page. If the velocity is perpendicular to the magnetic field lines, then boththe velocity and the magnetic force are parallel to the page, as in Figure 6.

Figure 6In this case, the particle is negatively charged, with the magneticfield directed into the page, perpendicular to the velocity. Todetermine the direction of the magnetic force, point your thumb inthe opposite direction of the velocity because the charge isnegative.

Example 1)An electron accelerates from rest in a horizontally directed electric fieldthrough a potential difference of 46 V. The electron then leaves theelectric field, entering a magnetic field of magnitude 0.20 T directed intothe page (Figure 7).(a) Calculate the initial speed of the electron upon entering themagnetic field. 4.0 x 106m/s

(b) Calculate the magnitude and direction of the magnetic force on theelectron. 1.3 x 10�13 N.

(c) Calculate the radius of the electron�s circular path. 1.1 x 10�4 m

Example 2) Calculate the mass of chlorine-35 ions, of charge 1.60 x 10�19 C, accelerated into amass spectrometer through a potential difference of 2.50 x 102 V into a uniform 1.00-Tmagnetic field. The radius of the curved path is 1.35 cm. 5.83 x 10�26 kg.

Understanding Concept.

1) Determine the magnitude and direction of the magnetic force on a proton moving horizontallynorthward at 8.6 x 104 m/s, as it enters a magnetic field of 1.2 T directed vertically upward.(The mass of a proton is 1.67x 10�27 kg.) 1.7 x 10�14 N [E]

2) An electron moving through a uniform magnetic field with a velocity of 2.0 x 106 m/s [up]experiences a maximum magnetic force of 5.1x 10�14 N [left]. Calculate the magnitude anddirection of the magnetic field. 0.16 T [horizontal, toward observer]

3) Calculate the radius of the path taken by an α particle (He2+ ion, of charge 3.2 x 10�19 C andmass 6.7x 10�27 kg) injected at a speed of 1.5 x 107 m/s into a uniform magnetic field of 2.4 T,at right angles to the field. 0.13 m

4) Calculate the speed of a proton, moving in a circular path of radius 8.0 cm, in a planeperpendicular to a uniform 1.5-T magnetic field. What voltage would be required toaccelerate the proton from rest, in a vacuum, to this speed? (m proton = 1.67 x 10�27 kg)1.1 x 107 m/s; 6.9 x 105 V

5) An airplane flying through Earth�s magnetic field at a speed of 2.0 x 102m/s acquires a chargeof 1.0 x 102C. Calculate the maximum magnitude of the magnetic force on it in a regionwhere the magnitude of Earth�s magnetic field is 5.0 x 10�5 T. 1.0 N