Chapters 29 and 30

Level I:

Magnetic Field:

Magnetism is the result of electric charge motion.  As soon as an electric charge moves, it creates a magnetic effect perpendicular to its direction of motion.  This can be easily verified by the a magnetized nail as shown.  If a steel nail is placed inside a coil of copper wire that is connected to a battery, the nail becomes a magnet.  Since the nail itself is perpendicular to the plane of each  loop of the coil; therefore, we can say that the current, or the motion of electric charges is perpendicular to the nail.  See the figure shown below:

Fig. 1

The reason for naming the poles of the nail as N and S is that if a bar magnet is hung at its middle by a string, it turns and aligns itself  toward almost the North and South poles of the Earth.  The end pointing to the North is called its North pole and the other end pointing to the South is called the South pole of the magnet.  The Geographic North and the Magnetic North are off by a few degrees however, and the position of the magnetic north changes slightly over years.  The reason for the magnetic effect of the Earth is also the motion of charged particles.  The molten metal at the core of the Earth is ionized and has a rotational motion parallel to the Equator creating a huge magnet which field must be perpendicular to the Equator plane and therefore passes through the North pole and the South pole of the Earth.  If the magnetic north and south are not exactly where the geographic north and south are, it is because the plane of rotation of the Earth's core (molten material) is not exactly parallel to the Equator.

Magnetic Field Lines:

Magnetic field lines are generally accepted to emerge from the North pole of a magnet and enter its South pole as shown.  This can be easily verified by placing a tiny compass around a bar-magnet at different positions.

Fig. 2

It is impossible to separate the North and South poles of a magnet.  They coexist.  If a bar magnet is cut at the middle (its neutral line), each piece becomes an independent magnet possessing a South pole and a North pole.  This is not the case with electric charges.  It is possible to have a separate negative charge and a separate positive charge. 

The Theory Behind Magnetism:

As we know, atoms are made of negative electrons, positive protons, and neutral neutrons.  Protons and neutrons have almost the same mass but are much heavier than electrons.   Protons and neutrons form the nucleus.  Electrons orbit the nucleus.  Electrons are considered the moving charges in atoms.  Electrons generate magnetic fields perpendicular to their planes of rotation.  The following argument gives you an idea of how some materials exhibit magnetic property although this does not reflect the exact picture.

Visualize a single electron orbiting the nucleus of its atom.  For simplicity, visualize a sphere in which this electron spins at a rate of say 1015 rotations per second.  This means one thousand trillion turns every second.  Therefore, at every one-thousand-trillionth of a second it possesses a particular plane of rotation in space.  In other words, the orientation of its plane of rotation changes 1015 times every second.  That is why we say it creates an electron cloud.  Only three of such orientations are sketched below.   For each plane of rotation, the magnetic field vector is shown to have its maximum effect at the center of the circle of rotation and perpendicular to that circle.

Fig. 3

An object of mass 1kg, for example, contains a very large number of atoms, and each atom depending on the element contains several electrons and each electron at any given instant of time has its own orientation of rotation and its own direction of magnetic field vector.  We are talking about hundreds of trillion trillions different magnetic field vectors in a piece of material.  There is no guarantee that all such electrons have their magnetic field vectors oriented in a single direction so that their magnetic effects add up.  An orbital  is a space around a nucleus where the possibility of finding electron is high.  An orbital can be spherical, dumbbell-shaped, or of a few other geometric shapes.  We assumed a spherical orbital for simplicity.  Each orbital can be filled with 2 electrons.  The two electrons in each orbital have opposite rotation directions.   This causes their magnetic field vectors have opposite directions as well.  The result is a zero net magnetic effect. This way, each atom that contains an even number of electrons will have all of its orbitals filled with pairs of electrons.  Such atoms are magnetically neutral.  However, atoms that contain odd numbers of electrons will have an orbital that is left with a single electron.   Such atoms are not magnetically neutral by themselves.  They become magnetically neutral, when they form molecules with the same or other atoms.  There are a few elements such as iron, cobalt, and nickel that have a particular atomic structure. This particular structure allows some outer shells orbitals to have single (unpaired) electrons.  Under normal circumstances, there is no guarantee that all orbitals of the atoms in a piece of iron, for example, to have their magnetic field vectors lined up parallel to each other.  But if a piece of pure iron is placed in an external magnetic field, the planes of rotation of those single electrons line up such that their magnetic fields line up with the direction of the external field, and after the external field is removed, they tend to keep the new orientation and therefore the piece of pure iron becomes a magnet itself.  The overall conclusion however is the fact that magnetism is the result of electric charge motion and that the magnetic field vector is perpendicular to the plane of rotation of oriented charged particles (mostly electrons).

Like poles and Unlike poles:

Like poles of two magnets repel and their unlike poles attract.  This is similar to electric charges.  Recall that like charges repel and unlike charges attract.  One difference is that separate positive and negative charges are possible, but separate magnetic North poles and South poles are not.  See Fig. 4.

Fig. 4

 

Uniform Magnetic Fields:

The magnetic field lines around a bar magnet is not uniform and varies with distance from its poles, Fig. 2.  The reason is that the field lines around a bar magnet are not parallel.  The density of field lines (number of lines per unit volume, say per cm3, for example) is a function of distance from the poles.  In order to make parallel magnetic field lines, the bar magnet must be bent into the shape of a horseshoe.  The field lines that emerge from the N-pole of the magnet then have to directly enter its S-pole and become necessarily parallel.  This is true only for the space in between the poles, Fig. 5.

Fig. 5

 

The Force of a Magnetic Field on a Moving Charge:

When a moving charge enters a magnetic field such that field lines are crossed, the charge finds itself under a force perpendicular to its direction of motion that gives it a circular motion.  When a charge enters a magnetic field such that its motion direction stays parallel to the field lines and no field line is crossed, then the charge will not be affected by the field and keeps going straight.   Imagine you are in a classroom facing the board and visualize a downward uniform magnetic field B similar to a rainfall.   Visualize a positive charge q entering horizontally from the left side going to the right side at speed v.  This positive charge will initially be forced toward the board as shown in Fig. (b).  Of course, the horizontal motion of the charge at velocity v will be perpendicular to the downward B.    Fig. (a) shows a general scenario in which q enters slanted where V is not normal to B.  In that case q will still be pushed toward the board, but with a weaker force.  Each subsequent field line that q crosses will change its direction slightly to where q gets caught in the magnetic field and B forces it to take on a circular path.  If we look from the top down onto q, we will see it spin counterclockwise.

In general, With B downward, and q moving to the right at velocity v,  the magnitude of F that initially pushes q  toward the board is given by:

F = q v B sin θ       

where θ is the angle between vectors B and V.

If θ = 90o, then  F = q v B.

q can be (+) or (-).   V can be to the left or right.   B can be upward or downward.   If any one of these 3 factors reverses, the direction of F reverses.  Now, if two of the factors reverse,  the direction of F remains the same.   If all 3 factors reverse, the direction of F reverses as well. 

The SI unit for magnetic field strength, B, is "Tesla"  shown by T. 

1Tesla (1T) is the strength of a magnetic field that exerts 1N of force on1C charge if the charge is crossing it at a speed of 1m/s normal to its lines.

 

Example 1:  A 14-μC charge enters a 0.030T magnetic field at a speed of 1.8x105m/s.   Find the magnitude of the force on it.

Solution:  F = q v B  ;  F = ( 14-μC )( 1.8x105 m/s)( 0.030 T) =  0.076N.

 

Example 2:  A 14-μC charge moving at 1.8x105 m/s enters a 0.030T downward magnetic field from the left side.    Its trajectory makes a 65o angle with respect to the field lines.   Find the magnitude and direction of the initial force on it as soon as it crosses a field line.  Refer to Fig. (a) above.

Solution: Referring to Fig. (a), it is clear that the charge will initially be pushed away from us (the observers).  The magnitude of this initial push is

F = q v Bsin θ    ;    F = ( 14-μC )( 1.8x105 m/s)( 0.030 T) sin (65o )=  0.069N.

 

Example 3:  An electron enters a 0.013-T magnetic field normal to its field lines and experiences a 3.8x10-15N force.  Determine (a) its speed and (b) its acceleration. The mass and charge of electron are:  Me = 9.108x10-31kg  and  q = |e-| = 1.6x10-19Coul, respectively.  The charge of electron is (-), but here we just need to use its absolute value.

Solution: F = q v B    ;    v = (F / qB) = ( 3.8x10-15N) / [(1.6x10-19C)(0.013T)]  =  1.8x106 m/s.

                F = Ma  ;   a = F / M =   3.8x10-15N / 9.108x10-31kg   = 4.2x1015 m/s2.

 

Motion of a charged Particle in a Magnetic Field:

So far, we have learned that when a charged particle crosses magnetic field lines, it is forced to change direction.  This change of direction does not stop as long as there are field lines to be crossed in the pathway of motion of the charged particle.  Magnetic field lines keep changing the direction of motion of the charged particle and if the field is constant in magnitude and direction, it gives a circular motion to the charged particle.  The reason is that F is perpendicular to V at any position inside the field as long as there are field lines to be crossed. This defines the concept of centripetal force.  Recall the concept of centripetal force.  Centripetal force, Fc,  is always directed toward the center of rotation.  Such force makes an object of mass M traveling at speed V to go around a circle of radius R.  In fact it is the force of magnetic field, Fm, that provides the necessary centripetal force, Fc.  We may equate the two after comparing the two figures below:

This formula is useful in finding the radius of rotation of a charged particle caught in a magnetic field.

 

Example 4:  A proton ( q = e+ = 1.6x10-19C,  M =  1.67x10-27kg) is captured in a 0.107T magnetic field and spins along a circle of radius 4.5 cm.  Find its speed knowing that it moves perpendicular to the field lines.

Solution:  R = (Mv) / (qB)  ;  solving for v:   

                v = RqB / ;   v = (0.045m)(1.6x10-19C )(0.107T) / 1.67x10-27kg  =  4.6x105 m/s.

 

Example 5:  In a certain device, alpha-particles enter a 0.88-T magnetic field perpendicular to its field lines.  Find the radius of rotation they attain if each carries an average kinetic energy of 520 keV.  An alpha-particle is a helium nucleus.  It contains 2 protons and 2 neutrons.  Mp = 1.672x10-27kg  and  Mn = 1.674x10-27kg.

Solution: Since the K.E. of each alpha-particle is given, knowing its mass ( 2Mp + 2Mn ), its speed can be  calculated.   K.E. = (1/2)Mv2.   Note that 1eV = 1.6x10-19J ; therefore,  1keV = 1.6x10-16J.

K.E. = (1/2)Mv2    ;    520(1.6x10-16J) = (1/2)[ 2(1.672x10-27kg) + 2(1.674x10-27kg) ] v2.

v = 5.0x106 m/s    ;    R = (Mv) / (qB)    ;  

 R = [ 2(1.672x10-27kg) + 2(1.674x10-27kg) ](5.0x106 m/s) / [(2 x 1.6x10-19C)( 0.88T)] = 0.12m = 12cm.

Note that each alpha-particle has 2 protons and carries 2 x 1.6x10-19C of electric charge.

 

Test Yourself 1:

1) In magnetizing a nail that is wrapped around with a coil of wire, the direction of the electric current in the loops of the wire is (a) parallel to the nail   (b) perpendicular to the nail if the loops are closely packed   (c) almost perpendicular to the nail if the loops are not closely packed  (d) b & c.      click here.

2) The direction of the magnetic field in a magnetized nail is (a) along the nail  (b) perpendicular to the nail  (c) neither a nor b.

3) If the four bent fingers of the right hand point in the direction of current in the loops of a magnetized coil, then the thumb points to (a) the South pole of the magnet  (b) the North pole of the magnet  (c) the direction normal to the magnet.      click here.

4) The magnetized nail experiment shows that (a) magnetic field occurs anywhere that there is an iron core  (b) anywhere a charged particle moves, magnetic effect develops in all directions  (c) anywhere a charged particle moves, there appears a magnetic effect that is normal to the direction of the charged particle's motion.

5) An electron orbiting the nucleus of an atom (a) does not develop a magnetic field because its radius of rotation is extremely small  (b) generates a magnetic effect that is of course normal to its plane of rotation at any instant  (c) cannot generate any magnetic effect because of its extremely small charge.      click here.

6) In a hydrogen molecule, H2 , the net magnetic effect caused by the rotation of its two electrons is zero because (a) at any instant, the two electrons spin in opposite directions creating opposite magnetic effects  (b) the instant its two electrons pass by each other, they repel and change planes of rotation that are opposite to each other causing opposite magnetic effects.  (c) both a & b.

7) The reason that atoms, in general, are magnetically neutral is that (a) electrons of atoms exist in pairs spinning in opposite directions in every orbital thereby neutralizing each other's magnetic effect.  (a) True  (b) False     click here.

8) The reason iron and a few other elements can maintain magnetism in them is that (a) these elements can have orbits in them that contain unpaired electrons  (b) under an external magnetic field, the orbits in these element with a single electron in them can orient themselves to the direction of the external field and stay that way  (c) both a & b.

9) For a bar-magnet, the magnetic field lines (a) emerge from its South pole and enter its North pole  (b) emerge from its North pole and enter its South pole  (c) emerge from its poles and enter its middle, the neutral zone.      click here.

10) If a bar magnet is cut at its middle, the neutral zone, (a) one piece becomes a pure North pole and the other piece a pure South pole  (b) both pieces will have their own South and North poles because magnetic poles coexist  (c) neither a nor b.

11) The magnetic field strength around a bar magnet is (a) uniform  (b) nonuniform and therefore varies with distance from its poles  (c) uniform at points far from the poles.      click here.

12)  The magnetic field in between the poles of a horseshoe magnet is (a) uniform  (b) nonuniform  (c) zero.

13) The magnetic field in between the poles of a horseshoe magnet (a) varies with distance from its either pole  (b) is directed from N to S  (c) has a constant magnitude and direction and is therefore uniform.  (d) b & c.      click here.

Problem: Visualize you are sitting in a class facing the board.  Suppose that the ceiling is the North pole of a huge horseshoe magnet and the floor is its South pole; therefore, you are sitting inside a uniform downward magnetic field.  Also visualize a fast moving positive charge emerging from the left wall and is heading toward the right wall; in other words, the velocity vector of the positive charge acts to the right.  Answer questions 14 through 18:

14) The charge will initially be pushed (a) toward you    (b) downward    (c)  toward the board.

15) The charge will take a path that is (a) straight toward the board  (b) circular at a certain radius of rotation  (c) curved upward.      click here.

16) If the radius of curvature is small such that the charge does not leave the space between the poles of the magnet it will have a circular motion that looking from the top will be (a) counterclockwise  (b) clockwise  (c) oscillatory.

17) If instead, a negative charge entered from the left side, looking from the top, it would spin (a) counterclockwise  (b) clockwise.

18) If a positive charge entered from the right side heading for the left, looking from the top again, it would spin (a) clockwise   (b) counterclockwise.      click here.

19) If the polarity of a magnetic field is reversed, the spin direction of a moving charged particle caught in it will (a) remain the same  (b) reverse as well.

20) The force F of a magnetic field B on a moving charge q is proportional to the (a) filed strength B   (b) particle's velocity V  (c) the amount of the charge q  (d)  sinθ of the angle that vector V makes with vector B  (e) a, b, c, & d.

21) The force, F of a magnetic field, B on a moving charge, q is given by (a) F = qB  (b)  F = qV  (c) F = qvBsinθ.

22) In the formula F = qvBsinθ, if q is in Coulombs, v in m/s, and F in Newtons, then B is in (a) N/(Coul. m/s)  (b) Tesla  (c) a & b.      click here.

23) The magnitude of F that a 0.0025T magnetic field exerts on a proton that enters it normal to its field lines at a speed of 3.7x106 m/s is (a) 1.5x1015N    (b) 0   (c) 1.5x10-15N.

24) An electron moves at a speed of 7.4x107m/s parallel to a uniform magnetic field.  The field force of on the electron is (a) 3.2x10-19N    (b) 0    (c) 4.8x10-19N.   click here.

25) The force that keeps a particle in circular motion is (a) circular force   (b) centripetal force   (c) tangential force.

26) When a charged particle is caught in a magnetic field and it keeps spinning at a certain radius of rotation, the necessary centripetal force is (a) the force of the magnetic field on it   (b) the electric force of the charged particle itself   (c) both a & b.      click here.

27) Equating the force of magnetic field, Fm , and the centripetal force, Fc , is like  (a) Mv/R = qvB     (b) v2/R = qvB      (c) Mv2/R = qvB.

28) In the previous question, solving for R yields: (a) R = qvB/(Mv2)    (b) R = Mv /(qB)     (c) both a & b.

29) The radius of rotation that a 4.0-μC charge carried by a 3.4-μg mass moving at 360 m/s normal to a 0.78T magnetic field attains is (a) 39cm   (b) 3.9m    (c) 7.8m.    click here.

30) One electron-volt of energy (1eV) is the energy of (1 electron) in an electric field where the potential is 1 Volt.  This follows the formula (a) Ue = qV where q is replaced by the charge of 1 electron and V is 1volt   (b) P.E. = Mgh   (c) neither a nor b.

31) Knowing that 1eV = 1.6x10-19J, if a moving proton has an energy of 25000eV, its energy calculated in Joules is (a) 4.0x10-15J    (b) 1.56x1023 J    (c) 4.0x10-19J.      click here.

32) A 25-keV proton enters a 0.014T magnetic field normal to its field lines.  Mp = 1.672x10-27kg.  The radius of rotation it attains is (a) 1.63m    (b) 2.63m    (c) 3.63m.

 

Velocity Selector:

It is possible to run a charged particle through a magnetic field perpendicular to the field lines along a straight path.  All one has to do is to place an electric field in a way that neutralizes the effect of the magnetic field.  Let's visualize sitting in a classroom (facing the board of course) in which the ceiling is the N-pole, floor the S-pole, and a positive charge is to travel from left to right horizontally that means normal to the downward magnetic field lines.  As was discussed before, the magnetic field does initially push the positive charge toward the board.  Now, if the board is positively charge and the back wall negatively charged, the charged particle will be pushed toward the back wall by this electric field.  It is possible to adjust the strengths of the magnetic and electric fields such that the forces they exert on the charge are equal in magnitude but opposite in direction.  This makes the charge travel straight to the right without deviation.  The resulting apparatus is called a "velocity selector."   For a velocity selector we may set the magnetic force on the charge equal to the electric force on the charge:   Fm = Fe.    This results in

q v B = q E        ;        v B = E ,       or        v = E / B.

If there is a large number of charged particles traveling at different speeds but in the same direction, and we want to separate the ones with a certain speed from the rest, this device proves useful.

Example 6:  In a left-to-right flow of alpha-rays (Helium nuclei = 2p+2n) coming out of a radioactive substance a 0.0225T magnetic field is placed in the downward direction.  What magnitude electric field should be placed normal to B such that only 0.050MeV alpha-particles survive both fields and go straight through?

Solution:   K.E. = 0.050 MeV means 0.050 mega electron-volts that means 0.05x106eV. 

Since each eV is equal to 1.6x10-19 Joules ; therefore,  K.E. = 0.050 x 106 x 1.6x10-19 J  = 8.0x10-15J.

To find v from K.E., use K.E. = (1/2)Mv2.  Using M = 6.692x10-27kg (verify) for the mass of an alpha-particle, the speed v is   v = 1.5x106 m/s.

Using the velocity selector formula:  v = E / B  ;  E = vB   ;   E = (1.5x106 m/s)(0.0225T) = 34000 N/C.

 

Cyclotron:

An application of the foregoing discussion is in the "cyclotron."  A cyclotron is a device that accelerates charged particles for nuclear experiments.  It works on the basis of the motion of charged particles in electric and magnetic fields.  When a particle of mass M and charge q moving at velocity v is caught in a magnetic field B, as we know, it takes a circular path of radius R given by

R = Mv/(qB).

The space in which such particles spin is cylindrical.  To accelerate the spinning particles to higher and higher velocities, the cylinder is divided into two semi-cylinders called the "Dees" as shown below.  The dees are connected to an alternating voltage.  This makes the polarity of the dees alternate at a certain frequency.   It is arranged such that when positive particles are in one of the dees, that dee becomes positive to repel those positive particles and the other dee is negative to attract them and as soon as the particles enter the negative dee, the polarity changes, the negative dee becomes positive to repel them again.  This continual process keeps accelerating the particles to a desired speed.  Of course, as the speed changes, the particles acquire greater and greater radii of rotation to where they are ready to leave the cylindrical space at which point they bombard the target nuclei under experiment.  A sketch is shown below:

If the speed of the particles become comparable to the speed of light (3.00x108 m/s in vacuum), their masses increase according to the Einstein's theory of relativity.  The mass increase must then be taken into account when calculating the period of rotation and energy of the particles.  These type of calculations are called the "relativistic" calculations.

Period of Rotation:

Period (T), the time it takes for a charged particle to travel one full circle or 2πR, can be calculated.  From the definition of linear speed, v = Δx/Δt , the linear speed of a particle in 1 full turn along a circular path is  V = 2πR /T  because T is the time of one rotation.  solving for T yields:

T = 2πR / v        (*)

v may be found from the formula for radius of rotation  R = Mv /(qB).  This yields:  v = qBR / M.  Substituting for v in (*), yields:

 T = 2πM / (qB)                

 

Example 7:  In a cyclotron, protons are to be accelerated.  The strength of the magnetic field is 0.024T.  (a) Find the period of rotation of the protons, (b) their frequency, (c) their final speed if the final radius before hitting the target is 2.0m, and (d) their K.E. in Joules and eVs.

Solution:  (a) T = 2πM /(qB)  ; T = (2π x 1.672x10-27kg) /[(1.6x10-19C)(0.024T)=  2.7x10-6 s.

                (b) f = 1 / T    ;    f  = 3.7x105 s-1    or    f = 3.7x105 Hz.

                (c) v = Rω    ;    v = R (2πf)   ;  v = (2.0m)(2π)(3.7x105s-1) = 4.6x106 m/s.

This speed (although very high) is still small enough compared to the light speed ( 3.00x108 m/s ) that the relativistic effects can still be neglected.

                (d) K.E. = (1/2)Mv2     ;    K.E. = (1/2)(1.672x10-27kg)(4.6x106m/s)2 = 1.8x10-14 J.

                     K.E. = 1.8x10-14 J ( 1 eV / 1.6x10-19 J) = 110,000eV  = 110KeV  = 0.11 MeV.

An Easy Relativistic Calculation:

According to the Einstein's theory of relativity, when mass M travels close to speed of light it becomes more massive and more difficult to accelerate further.  The mass increase effect is given by the following formula:

Example 8:  In a cyclotron electrons are accelerated to a speed of 2.95x108 m/s.  (a) By what factor does the electron mass increase?  Knowing that the rest mass of electron Mo = 9.108x10-31kg, determine (b) its mass at that speed.

Solution:  (a)  Let's find γ step by step.  Let's first find (v/c), then (v/c)2 that is the same as v2/c2, then 1- v2/c2, then the square root of it, and finally 1 over that square root.  The sequence is as follows:

(v / c) = 2.95 / 3.00 = 0.98333...    (Note that 108 powers cancel).

v2 / c2    =   (v / c)2   =   (0.98333...)2  =   0.96694...

1 - v2 / c2   =   1 - 0.96694...   =   0.0330555...

SQRT( 1 - v2 / c2 )  =   0.1818119

γ  =  1 / SQRT( 1 - v2 / c2 )    =     1 / 0.1818119     =   5.50    (The # of times mass of electron increases)

(b) M = Moγ         ;        M  =  ( 9.108x10-31kg )( 5.50 )  =  5.01x10-30 kg.

 

Example 9:  Find the value of γ for the protons in Example 7.

Solution:  To be done by students 

 

Sources of Magnetic Field:

Aside from permanent magnets, magnetic fields are mostly generated by coils of wire.  A coil is a wire that most often is wrapped around a cylinder.  Most coils are cylindrical.  Long coils produce a fairly uniform magnetic field inside them specially toward their middle and along their axis of symmetry.  To understand the magnetic field inside a coil, we need to know the magnetic field around a long straight wire as well as that of a single circular loop.

Magnetic Field Around a Straight and Long Wire:

For a very long and straight wire carrying a current I, we expect the magnetic field B to be perpendicular to the direction of the current.  Since anywhere around the wire this property must equally exist, the magnetic field lines are necessarily concentric circles with the current (the wire) perpendicular to the planes of the circles at their shared center.  The figure is shown below.

As r increases, B of course decreases as is also apparent from the equation shown above.  The direction of B is determined by the right-hand rule again.  If the thumb shows the direction of I, the four bent fingers point in the direction of B.  In the above formula μo = 4π x 10-7 Tm/A is called the "permeability of free space (vacuum)" for the passage of magnetic field lines.  For any material or substance permeability μ may be measured.   For every material a constant may then be defined that relates μ to μo.  

Example 10:  If in the above figure, I = 8.50A, determine the magnitude of B at r = 10.0cm, 20.0cm, and 30.0cm.

Solution:  Using the formula B = μoI / 2πr, we get: 

B1 = 1.7x10-5 Tesla   ;     B2 = 8.5x10-6 Tesla    ;      B3 = 5.7x10-6 T.

 

Magnetic Field of a Current-Carrying Circular Loop:

The magnetic field produced by a current-carrying circular loop is necessarily perpendicular to the plane of the loop.   The reason is that B must be perpendicular to the current I that the loop carries.  The direction is determined by the right-hand rule as was discussed in the nail example at the beginning of this chapter.  The magnitude of B at the center of the loop is maximum and given by the following formula.  Pay attention to the figure as well.

Example 11:  If in the above figure, I = 6.80Amps, determine the magnitude of B for r = 10.0cm, 20.0cm, and 30.0cm.

Solution:  Using the formula B = μoI / 2r, we get: 

   B1 = 4.3x10-5 T   ;    B2 = 2.1x10-5 T    ;    B3 = 1.4x10-5 T.

 

Magnetic Field Inside a Solenoid:

A solenoid is a long coil of wire for which the length-to-radius ratio is not under about 10.  The magnetic field of a single loop of wire is weak.  A solenoid has many loops and the field lines inside it specially in the vicinity of its middle are fairly parallel and provide a uniform and stronger field.  Placing an iron core inside the solenoid makes the field even stronger, some 400 times stronger.   μiron = 400μo.  (See figure below).  The formula for magnetic field strength of a solenoid is:

B = μonI    where n is the number of loops per unit length.

n in SI units is # of loops per meter of the solenoid.

Example 11:  A solenoid is 8.0cm long and has 2400 turns.  A 1.2-A current flows through it.  Find (a) the strength of the magnetic field inside it toward its middle.  (b) If an iron core is inserted in the solenoid, what will the field strength be?

Solution: (a) B = μonI    ;    B = (4π x 10-7 Tm/A)(2400turn / 0.080m) )(1.2A ) = 0.045T.

                (b)  Iron increases μo by a factor of 400; therefore,   μ = (400)(0.045T) = 20T.

One Application of Solenoid:

Anytime you start you car, a solenoid similar to the one in the above example gets magnetized and pulls an iron rod in.  The strong magnetic field of the solenoid exerts a strong force on the iron rod (core) and gives it a great acceleration and high speed within a short distance.  The rod is partially in the solenoid to begin with and gets fully pulled in after the solenoid is connected to the battery the ignition key is turned to crank the engine.  The current that feeds the solenoid may not be even 1A, but the connection it causes between battery and starter pulls several amps from the battery.  The forceful moving rod inside the solenoid collides with a copper connector that connects the starter to the battery.  This connection allows a current of 30Amps to 80Amps to flow through the starter motor and crank the vehicle.  The variation of the amperage depends also on how cold the engine is.  The colder the engine, the less viscous the oil, and the more power is needed to turn the crankshaft.

Example 12: The magnetic field inside a 16.3cm long solenoid is 0.027T when a current of  368 mA flows through it.  How many turns does it have?

Solution:  B = μonI  ;  n = B /(μoI)  ;   n =  0.027T / [(4π x 10-7 Tm/A)(0.368A )] = 58400 turns /m.

This is the number of turns per meter.  If the solenoid was 1.00m long, it would have 58400 turns.  It is only 0.163m long, and therefore it has less number of turns.  If N is the number of turns, we may write:

N = nL    ;    N = (58400 turns / m)(0.163m) = 9520 turns.

Definition of Ampere:

We defined 1A to be the flow of 1C of electric charge in 1s.  A preferred definition for the unit of electric current or Ampere is made by using the force per unit length that two infinitely long parallel wires exert on each other.  Recall that when an electric current flows through an infinitely long and straight wire, it generates a magnetic field around it that can be sensed along concentric circles perpendicular to the wire.  If two of such wires are parallel to each other and the current in them flow in the same direction, they attract each other.  If the current flows in them in opposite directions, they repel each other.  The magnitude of the force they exert on each other depends on the distance between the wires and the currents that flow through them. 

If two parallel wires that are 1m apart have equal currents flowing in them in the same direction, and the two wires attract each other with a force of 10-7N/m in vacuum, then the current through each wire is  1Amp.

 

Test Yourself 2:

1) A velocity selector takes advantage of (a) two perpendicular electric fields  (b) a set of perpendicular electric and magnetic fields  (c) two perpendicular magnetic fields.    click here.

2) The forces (Fm and Fe) that the magnetic and electric fields of a velocity selector exert on a charge q, must be (a) equal in magnitude  (b) opposite in direction  (c) both a & b.

3) Fm and Fe in a velocity selector are given by (a) Fm = qvB and  Fe = qE     (b) Fm = qB and  Fe = qE    (c) Fm=qvB and  Fe=qE  such that Fm = Fe.      click here.

4) Setting Fm = Fe and solving for v results in (a) V = B/E    (b) V = E/B    (c) V =EB.

5) The formula V = E / B (a) depends on the amount of charge   (b) does not depend on the amount of the charge   (c) does not depend on the sign of the charge   (d) b & c.

6) What strength uniform electric field must be placed normal to a 0.0033T uniform magnetic field such that only charged particles at a speed of 2.4x106m/s get passed through along straight lines?  (a) 7900N/C     (b) 9700N/C   (c) 200N/C.      click here.

7) A cyclotron is a device that is used to (a) accelerate charged particles to high speeds and energies   (b) accelerate charged particles to speeds close to that of light   (c) perform experiments with the nuclei of atoms   (d) a, b, & c.

8) Speed of light is (a) 3.00x108m/s      (b) 3.00x10-8m/s      (c) 3.00x105 km/s    (d) a & c.

9) A speed of 3.00x10-8m/s is (a) faster than speed of light     (b) slower than the motion of an ant   (c) even germs may not move that slow   (d) extremely slow, almost motionless   (e) b, c, and d.      click here.

10) In a cyclotron, a charged particle released near the center (a) finds itself in a perpendicular magnetic field and starts spinning    (b)  spins at a certain period of rotation given by T = 2πM/(qB)   (c) is also under an accelerating electric field that alternates based on the period of rotation of the charged particle    (d) a, b, and c.

11) As the particles in a cyclotron accelerate to high speeds comparable to that of light (a) a mass increase must be taken into account   (b) the mass increase affects the period of rotation    (c) a & b.

12) The magnetic field around a current-carrying long wire (a) is perpendicular to the wire, and at equal distances from the wire has the same magnitude   (b) is parallel to the wire   (c) both a and b.      click here.

13) The magnetic field around a current-carrying long wire (a) may be pictured as concentric circles at which the field vectors act radially outward.  (b)  may be pictured as concentric circles at which the field vectors act tangent to the circles.  (c) has a constant magnitude that does not vary with distance from the wire.  (d) b & c.

14) The formula for the field strength around a current carrying long wire is (a) B = μoI / (2πR)     (b) B = μoI / (2R)     (c) B = I / (2R)       click here.

15) μo= 4π x 10-7 Tm/Amp is called (a) the permittivity of free space for the passage of electric field effect.  (b) the permeability of free space for the passage of the magnetic field effect.  (c) neither a nor b.

16) The farther from a straight and long current-carrying wire, the (a) stronger the magnetic effect   (b) the more constant the magnetic effect   (c) the weaker the magnetic effect.      click here

Problem: Draw two concentric circles in a plane perpendicular to a wire that passes through the center of the circles.  Suppose that the wire carries a constant electric current, I, upward.  Also suppose that the radius of the greater circle is exactly twice that of the smaller circle.  You also know that if you were to show magnetic field vectors, you would draw them tangent to those circles.  Draw a vector of length say 1/2 inch tangent to the greater circle as the magnitude of B at that radius.  Then draw another vector tangent to the smaller circle to represent the field strength at the other radius.  Answer the following questions:

17) The magnitude of the field strength at the smaller radius is (a) a vector of length 1 inch    (b) a vector of length 1/4 inch   (c) a vector of length 1/16 inch.      click here.

18) Based on the upward current in the wire, and looking from the top, the direction of vectors you draw must be (a) clockwise    (b) counterclockwise.

19) What should be the length of the tangent vector you may draw at another circle which radius is 5 times that of the smaller circle?  (a) 1/25 inch     (b) 1/125 inch    (c) 1/5 inch.

20) The magnetic field that a current carrying loop of wire (circular) generates is (a) perpendicular to the plane of the loop   (b) has its maximum effect at the center of the loop and normal to it   (c) has an upward direction if the current flows in the circular loop horizontally in the counterclockwise direction   (d) a, b, & c.      click here.

21) A solenoid is a coil which length is (a)  at most 5 times its radius   (b) at least about 10 times its radius.

22) The magnetic field inside a solenoid and in the vicinity of its middle (a) is fairly uniform   (b) is non-uniform    (c) has  a magnitude of B = μon I where n is its number of turns   (d)  has  a magnitude of B = μon I where n is its number of turns per unit length   (e) a & d.

23) A solenoid is 14.0cm long and has 2800  loops.   A current of 5.00A flows through it.  The magnetic field strength inside and near its middle is (a) 0.0176T   (b) 0.126T     (c) 0.00126T.      click here.

24) The magnitude of B inside and at the middle of a 8.0cm solenoid is 0.377T.  It carries a 5.00A current.  The number of loops of the solenoid is (a) 4,800   (b) 12,000   (c) 6,000.

25) The magnetic field strength, B, at the center of a coil that has N loops and carries a current I is (a) B = NμoI /(2R)     (b) B = NμoI /(2πR).     click here.

Level II:

            Magnetic Force between Two Parallel and Current-carrying Wires:

The figure on the right shows two parallel and infinite wires that are a distance d apart and carry positive currents I1 and I2.   The magnetic field that I1 generates at the perpendicular distance d from it is labeled B1.   B1 is perpendicular to I2.   This makes the force of B1 on I2 to be directed toward wire 1 as labeled by F12.   (If you are facing the board in a classroom and B is downward, positive charges going from left to right will first be pushed toward the board).   Similarly, the force of B2 (caused by wire 2) on wire 1 labeled by F21 is toward wire 2 as shown.

The magnitude of F12 is:       F12 = I22B1,     (1)

and the magnitude of F21 is:   F21 = I11B2.    (2)

Since the two forces are toward each other, the wires attract each other.  For opposite currents they repel.

Note that 1 and 2 are treated as two vectors that show the direction of the moving charges in the wires.   If we choose each as a unit of length, say 1m in SI, the calculated forces will be per meter of the wires.

Since B1 = μoI1/(2πd), and B2 = μoI2/(2πd) (1) and (2) become

F12 = μoI1I22 /(2πd)   and  F21 = μoI1I21 /(2πd)  or, in general, force per unit length on each wire becomes:

F/ℓ = μoI1I2/(2πd)               (3)

            Definition of Ampere:

Based on Equation 3, if two infinite and parallel wires that are 1m apart carry equal currents and exert a force of 2x10-7N on every meter of each other, the current in each is 1A.

            Biot-Savart Law for a Current Element:

Biot and Savart along with assistance from mathematician Laplace, figured out a way to show some similarity between the magnetic filed due to an infinite current carrying wire and the electric field of an infinite line of electric charges.

The magnetic field generated by an infinite wire that carries current I is given by

B1 = μoI/(2πR)

If we write this as B1 = oI/(4πR) and let  k' = μo /(4π), then B1 becomes:

B1 = 2k' I/R               (4)

We do not have a static line of charge to just create a total electric field at P, for example.   We have a line of moving charges (Current I) that generates a total magnetic field at P.  Here, we use Id as a differential current element that creates a differential magnetic field dB at P.  The strongest dB belongs to that differential current Id that is just passing by C, the center of the circle shown.  Biot and Savart showed that the magnitude of dB as given below is the correct and valid form:

(5)

The current elements are of course continuous and are not isolated as was the case for a line of electric charges.

As θ increases and approaches 90o, The current element moves up and becomes closer to point P.  When θ is exactly 90o, Id is at C, and at its closest distance of R to point P.

            Example 13:

 

Find the magnetic field strength at a distance R from an infinite straight wire that carries a current I.

Solution:  We may use the above figure and add the contributions of all (Id)'s from -∞ to +∞.   To do this, it is better to use angle α.   Let α vary from (-α1) to (+α2).   Note that α and θ are complementary angles and sine of one equals cosine of the other.  We will replace sinθ by cosα.

Both d and r must be expressed in terns of R and α.   Since tanα = ℓ/R,  then ℓ = R tanα.  Differentiation results in dℓ=Rsec2α dα.

Since cosα = R/r, we get:  r = R /cosα  or  r = R secα.   Substituting for sinθ , r, and d, in dB, we get:

(6)

If α1 and α2 are replaced by /2 and π/2 respectively, B becomes:    B = μoI /(2πR) as expected.

 

            Example 14:

 

Calculate the magnetic field strength at a distance (y) from the center of a circular loop of radius (a) that carries a current (I).

Solution:  The current element Id is shown in the figure on the right.  For every Id there is an opposite direction Id on the opposite side of the loop.  Note that dB is normal to r.  In this figure every dB has a component along x and a component normal to x.  The components along x that means (dBx)'s add up, but the components normal to x that means (dBy)'s cancel due to symmetry.

Each component along x is  (dB)x = (dB)sinα.

 The main formula is the Biot-Savart formula that expresses dB in terms of Idℓ. We may write:

For points far from the center and still on the axis, let X   in which case the a2 of the denominator can be neglected and the result is:

            Example 15:

Calculate the magnetic field strength, B at a typical point along the axis of a solenoid of length  that contains N loops and carries a current I.

Solution: We need to add the contribution (dB)x of each loop at a given point along the axis.  Equation (*) above may be used.  It is easier to divide the solenoid into small packs of loops each with a length dx as shown.  If n = N/is the number of loops per unit length of the solenoid, then ndx is the number of loops within dx.  The differential current that ndx carries is therefore (ndx)I. 

From the figure x = a tanθ This makes dx = a sec2 θ dθ.

Replacing I in (*) by  nIdx = nIa sec2 θ   and  x by a tanθ, results in:

For an infinite solenoid, θ1 = -π/2 and θ2 = +π/2 that results in

B = μonI   as expected.

 

            Ampere's Law:

Ampere was not comfortable with the work of Biot and Savart.  He believed that the current element Id was not precise.  He wrote the experimental formula  B = μoI /(2πr) that we already know as

B(2πr) =  μoI

and stated that vector B is tangent to any circular path as shown in the figure and if its magnitude is multiplied by 2πr, it must be equal to μoI.  

Even for any arbitrary closed path that encloses the wire, if we calculate the tangential components of B along that arbitrary path, the result is equivalent to μoI.  The component of B along d is nothing but

  B dℓ = Bdℓ cosθ as shown.  θ is the angle between B and dℓ.

Integrating B d along the closed path around the wire can be set equal to μoI.  This is called the "Ampere's Law" written as:

where I is the current through the surface enclosed by the path.

The following figure shows an arbitrary path around the wire and how the tangential component of B may be calculated at each point along the path.

            Example 16:

Use Ampere's law to derive the formula for the magnetic field inside a solenoid.

Solution:  If we take a cross-sectional area of a solenoid that is cut in half along its axis, and choose a rectangular closed path that covers length L of loops, as shown, and apply the Ampere's law to the total current going through the closed path, B can be calculated along the axis of the solenoid.  Referring to the figure on the right, we may write the line integral as:

 

N = nL because

n = # of loops per meter of the solenoid.