Chapters 29 and 30
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. If a steel nail is placed inside a coil ( A coil is a piece copper wire wrapped around a cylinder) that is connected to a battery, the nail becomes a magnet. Since the nail itself is perpendicular to the surface of each loop of the coil; therefore, we can say that the current, or the motion of electric charges is perpendicular to the nail (magnetic field). See the figure shown below:
The reason for naming the poles of the nail as N and S is that if a 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 pole 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 whose field must be perpendicular to the Equator plane and therefore passes through the North pole and the South pole of the Earth.
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.
It is impossible to separate the North and South poles of a magnet. They coexist. If a bar magnet is cut from 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 it 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 turns 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 electronic cloud. 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.
An object of mass 1 lb. 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 orientation 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 electrons 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 directions. This causes the 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 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 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 electron.
Like poles and Unlike poles:
Like poles repel and 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 while a separate North poles and South poles are not. See the figure shown:
Uniform Magnetic Fields:
The magnetic field lines around a bar magnet is not uniform and varies with distance from its poles. The reason is that the field lines around a bar magnet are not parallel. The density of field lines 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. This is shown in the figure below.
Force of 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 the direction of motion that gives it a circular motion. If a charge enters a field such that its motion direction is 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. When you are in a classroom facing the board, visualize a downward uniform magnetic field (ceiling being the North pole and floor the South pole of the magnet). Visualize a positive charge entering from the left side of the classroom going toward the right side. This positive charge will initially be forced toward the board as shown:
If the downward field vector is (B), and the charge velocity crossing the magnetic field to the right and at right angle is (v), the magnitude of force (F) initially pushing the moving charge toward the board is given by:
F = q v B. Fig.2
If charge (q) is making an angle θ with the magnetic field lines, then (F) will have a smaller value given by:
F = q v B sin θ. Fig. 1
If (v) is to the left, (F) will be toward the class. Also if (B) is upward, F will be toward the class. If the charge is negative, (F) will be toward the class. Therefore, there are three elements that can affect the direction of (F). If any two of these three elements reverse simultaneously, the direction of (F) will remain the same.
The unit for B is expressed in Tesla (T). If (F) is in Newtons, (v) in m/s, and (q) in Coulombs, then, (B) will be in Tesla.
One Tesla of magnetic field strength is the strength that can exert a force of 1N on 1Coul. of electric charge that is moving at a speed of 1 m/s perpendicular to the magnetic field lines.
Example 1: A 14-μC charge enters from the left perpendicular to a downward magnetic field of strength 0.030 Tesla at a speed of 1.8x105 m/s. Find the magnitude and direction of the initial force on it as soon as it crosses a field line. Refer to Fig.2.
Solution: Referring to Fig. 2, it is clear that the charge will initially be pushed toward the board. The magnitude of this initial push is
F = q v B ; F = ( 14-μC )( 1.8x105 m/s)( 0.030 T) = 0.076N.
Example 2: A 14-μC charge enters from the left through a 65° angle with respect to a downward magnetic field of strength 0.030 Tesla at a speed of 1.8x105 m/s. Find the magnitude and direction of the initial force on it as soon as it crosses a field line. Refer to Fig.1.
Solution: Referring to Fig. 1, it is clear that the charge will initially be pushed toward the board. The magnitude of this initial push is
F = q v Bsin θ ; F = ( 14-μC )( 1.8x105 m/s)( 0.030 T) sin (65° )= 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 its speed.
Solution: F = q v B ; v = (F / qB) = ( 3.8x10-15N) / [(1.6x10-19C)(0.013T)] = 1.8x106 m/s
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 instant and position and that exactly 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 circule of radius R. In fact it is the force of magnetic field, Fm, that supplies 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 curvature of the circular path of a charged particle when caught in a magnetic field.
Example 4: A proton ( q = 1.6x10-19C, M = 1.67x10-27kg) is captured in a 0.107-T magnetic field an 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 / M ; 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)]
Each alpha-particle has 2 protons and carries 2 x 1.6x10-19C of electric charge. R = 0.12m = 12cm.
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 coil. (b) the North pole of the magnet coil. (c) the direction normal to the magnet coil. 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 must exist in pairs spinning in opposite directions thereby neutralizing each other's magnetic effect. (b) not all atoms are iron atoms and therefore do not have any magnetic effect in them. 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 that means 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 emerges from the left wall and is heading for the right wall; in other words, the velocity vector of the positive charge acts to the right. Answer the following questions:
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, 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 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 V makes with 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 the force that a 0.0025-T magnetic field exerts on a proton that enters it normal to its field lines and has 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 7.4x107 m/s parallel to a uniform magnetic field. The force that the magnetic field exerts on it 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 magnetic field on it that keeps it spinning. (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 a 4.0-μCoul. being carried by a 3.4-μg mass moving at 360 m/s normal to a 0.78-T magnetic field attains is (a) 39cm. (b) 3.9m (c) 7.8m click here
30) One electron-volt of energy (1 eV) is the energy of (1 electron) in an electric field where the potential is ( 1 Volt). This follows the formula (a) P.E. = 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 0.014-T magnetic field normal to its field lines. Each proton has a mass of 1.67x10-27kg. The radius of rotation if finds is (a) 1.63m (b) 2.63m (c) 3.63m.
It is possible to run a charged particle through a magnetic field perpendicular to the field lines without any deviations from 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 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) coming out of a radioactive substance a 0.0225-T magnetic field is placed in the downward direction. What magnitude electric field should be placed around the flow such that only 0.050MeV alpha-particles survive both fields?
Solution: K.E. = 0.050 MeV means 0.050 mega electron-volts that means 5.0x104 electron volts. Since each eV is equal to 1.6x10-19 Joules ; therefore, K.E. = 0.050 x 106 x 1.6x10-19 J ; K.E. = 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)(0.0225 T) = 34000 N/C.
An application of the foregoing discussion is in cyclotron. Cyclotron is a device that accelerates charged particles for nuclear experiments. It works on the basis of the motion of charged particles in 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 the particles spin is cylindrical. To accelerate the spinning particles to higher and higher velocities, the cylinder is divided into two semicylinders called the "Dees." 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 the 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 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), 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 circle or 2πR, can be calculated. From the definition of speed V = 2πR / T, 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 0.024 Tesla. (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 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 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 that 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 common 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 = 8.50Amps, 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 Tesla
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 at the center of the loop is 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 Tesla
B2 = 2.1x10-5 Tesla ; B3 = 1.4x10-5 Tesla
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 = 400 μo. (See figure below). The formula for magnetic field strength of a solenoid is:
B = μonI where n is the number of turns per unit length.
n in SI units is # of turns 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 the middle. (a) If an iron core is inserted in the solenoid, what will the field the field strength be?
Solution: (a) B = μonI ; B = (4π x 10-7 Tm/A)(2400turn / 0.080m) )(1.2A ) = 0.045 Tesla.
(b) Iron increases μo by a factor of 400; therefore, (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 in an iron rod. 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 by you when you try to crank the engine. The current that feeds the solenoid may not be even one amp, but the connection it causes between battery and starter pulls several amps from the battery. The forceful moving rod 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 your car. The variation of the amperage depends 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 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 0.0033-T uniform magnetic field such that only charged particles at a speed of 2.4x106m/s get passed through along straight lines? (a) 7900N/Coul. (b) 9700N/Coul. (c) 200N/Coul. 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 wire that carries a current, 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 whose 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 whose length is (a) at most 5 times its radius. (b) at least 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. turns. 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 magnetic field strength inside and at the middle of a 8.0cm long solenoid is 0.377T and it carries a 5.00-Amp current. The number of turns of the solenoid is (a) 4,800 turns. (b) 12,000 turns. (c) 6,000 turns.
25) The formula for the magnetic field strength, B, at the center of a coil, not a solenoid, than has N turns and carries a current, I, is (a) B = Nμo I / (2R). (b) B = Nμo I / (2πR). click here
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 a 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 = I2ℓ2B1, and
the magnitude of F21 is: F21 = I1ℓ1B2
Since the two forces are toward each other; therefore , the wires attract each other.
Note that ℓ1 and ℓ2 are treated as 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.
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 a 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 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 = I2ℓ2B1 (1)
The magnitude of F21 is: F21 = I1ℓ1B2 (2)
Since the two forces are toward each other, the wires attract each other. For opposite currents they repel.
Note that ℓ1 and ℓ2 are treated as 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 force will be force per meter of the wires.
Since B1 = μoI1/(2πd), and B2 = μoI2/(2πd), (1) and (2) become
F12 = μoI1I2ℓ2 /(2πd) and F21 = μoI1I2ℓ1 /(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 the same current and exert a force of 2x10-7N on every meter of each other, the current in each is 1 Amp.|
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 that an infinite wire carrying current I generates is given by
B1 = μoI/(2πR)
If we write this as B1 = 2μoI/(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 Idℓ as a differential current element that creates a differential magnetic field dB at P. The strongest dB belongs to that differential current Idℓ 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:
The current elements are of course continuous and not isolated as was the case for a line of electric charges.
As θ increases and approaches 90°, The current element moves up and becomes closer to point P. When θ is exactly 90°, Idℓ is at C, and at its closest distance of R to point P.
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 (Idℓ)'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 is equal to cosine of the other. We will replace sinθ by cosα .
Both dℓ and r must be expressed in terns of R and α . Since tanα = ℓ/R, then ℓ = Rtanα . Differentiation results in dℓ=Rsec2α dα .
Since cosα = R/r, we get r = R/cosα or r = Rsecα . Substituting for sinθ , r, and dℓ, in dB, we get:
If α1 and α2 are replaced by -π/2 and π/2 respectively, B becomes: B = μoI /(2πR) as expected.
|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 Idℓ is shown in the figure on the right. For every Idℓ there is an opposite direction Idℓ on the opposite side of the loop. In this figure every dB has a horizontal and a vertical component. The horizontal components (dBx) add up, but the vertical ones (dBy) cancel due to symmetry.
Each horizontal component 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 y → ∞ in which case the a2 of the denominator can be neglected and the result is:
|Calculate the magnetic field strength,
B at a typical point along the axis of a
solenoid of length
ℓ that contains
N loops and carries a current
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/ℓ 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 dθ 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 was not comfortable with the work of Biot and Savart. He objected that the idea of current element Idℓ was not precise. He wrote the experimental formula B = μoI /(2πr) for the magnetic field around an infinite current-carrying wire 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.
Ampere was not comfortable with the work of Biot and Savart. He objected that the idea of current element Idℓ 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 dℓ is nothing but B ∙dℓ = Bdℓ cosθ as shown.
Integrating B ∙dℓ along the closed path around the wire can be set equal to μoI. This establishes the Ampere's Law 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.
|Use Ampere's law to derive the formula for the 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 a length L of loops 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.