Chapter 31 and 32
When a charge moving a velocity V crosses magnetic field lines of intensity B, it experiences a force F perpendicular to the plane that contains V and B. This was discussed in Chapter 21 and can be easily observed by flowing a current through a straight wire placed in the field of a horseshoe magnet, as shown below:
In the left figure, key K is open, no current is flows through the loop (no charge moves at any velocity), and therefore, there is no force on the wire segment in the field. In the right figure, key K is closed, current I flows through the loop ( charge q moves at some velocity v), and therefore, force F is exerted on the wire segment in the magnetic field B.
The reverse process is also possible. If the battery is removed as shown in the left figure below, and the circuit is closed, nothing will happen. But if the loop is pushed forward with force F or swung forward, an electric current appears in the loop. This can be verified by placing a sensitive ammeter in the loop as shown below:
The conclusion is that when an electric current flows in a wire in a magnetic field, the field exerts a force on the wire and causes motion (electric motor), and when a wire or a loop is moved mechanically in a magnetic field, a current is generated (electric generator). This experiment is the basis for electric motors and generators. To understand the theory in better details, we need to learn the concept of magnetic flux.
If you hold a ring horizontally under rain, maximum number of rain drops pass through it (1). If you hold it vertically, no rain drop passes through it (4). If you hold it at some angle, some rain drops but not the maximum possible pass thorough it (2) and (3). See figures shown below. To show the orientation of the ring in space, it is better to measure the angle its normal makes with the vertical direction. Normal to the ring means the line that is perpendicular to its plane. In each case the flux of rain through the ring is different.
Magnetic field line may also be viewed as rain lines if we are talking about a downward magnetic field. The heavier the downpour, the stronger the magnetic field B.
Magnetic Flux Through a Surface:
The magnetic flux Φm of a magnetic field B through a surface of area A whose normal n makes an angle θ with the field lines, is defined as: Φm = B A cosθ An appropriate figure for this formula is shown below:
where symbol Φ is pronounced "phi". The SI unit for magnetic flux is Tm2 called " Weber." For a coil of wire that has N loops, the total flux is of course:
Φm = NB A cosθ
Example 1: A rectangular loop of wire (2.5cm X 4.0cm) is placed in between the poles of a horseshoe magnet such that the upward magnetic field lines make a 30.0˚ angle with the loop surface as shown. The magnetic field has a strength of 1.2 Tesla. Calculate the magnetic flux through the loop.
Solution: A = (0.025m)(0.040m) = 0.0010m2
The angle that B makes with the surface of the loop according to the problem is 30.0° as shown. θ , the angle that B makes with the normal to the surface can be calculated. θ = 90.0° - 30.0° = 60.0° and the flux is
Φm = BAcosθ =1.2T (0.001m2)cos(60.0°) = 6.0x10-4 Tm2
or, Φm = 6.0x10-4 Weber
Example 2: A student experimenting with magnets turns a circular loop of wire of radius 4.0cm from a horizontal position to a vertical position in between a 0.17-T horseshoe magnet. Calculate the change in magnetic flux through the circular loop.
Solution: A = π R2 = π ( .040m)2 = 0.0050 m2
When the loop is horizontal, θ = 0, and the flux is
Φ1 = B A cosθ = (.17 T)(0.0050m2)( 1 ) = 8.5x10-4 Tm2
When the loop is vertical, θ = 90, and the flux is
Φ2 = B A cosθ = (.17 T)(0.0050m2)( 0 ) = 0 Tm2
Therefore, the change in magnetic flux is:
Φ2 - Φ1 = 0 - 8.5x10-4 Tm2 = - 8.5x10-4 Tm2
Δ Φ = - 8.5x10-4 Tm2
Faraday's Law of Magnetic Induction:
When a coil of wire is moved toward one pole of a magnet or away from it, a current is induced in the coil that can be measured by a sensitive ammeter as shown in the figure below:
The key to the generation of electric current in a coil of wire is the change in the magnetic flux through the surface area of the loops of the coil. Faraday showed that the voltage generated across the terminals of a coil is directly proportional to the change in magnetic flux ΔΦm , and inversely proportional to the time interval Δt within which the flux change occurs. He came up with the formula:
In Faraday's formula, E is the voltage or electromotive force (emf) generated in the coil, N the number of loops of the coil, ΔΦm the change in the magnetic flux, and Δt is the change in time. In SI units, it is easy to show that E has the unit of volt as is expected. The ( - ) sign indicates that the direction of the induced (generated) current opposes the direction of the change in the flux that causes it. This will be discussed under Lenz's Law later.
Example 3: A coil has 750 turns and a radius of 6.0cm. A strong magnet is inserted in it such that in a time interval of 0.50 seconds, a change in magnetic field intensity of 1.36 T occurs. Find the induced voltage across the coil. The field lines stay perpendicular to the loops during the process.
Solution: ΔΦm = NBAcos( θ ) ; ΔΦm = (750)(1.36 T)(π)(0.060m)2 cos(0) = 11.5 Tm2
|E |= ΔΦm / Δt = 11.5 Tm2 / 0.50s = 23 volts
Example 4: A 125-loop coil of wire is in a 2.17-T magnetic field such that the field lines are parallel to its loops' surfaces. The area of each loop is 85.4 cm2. The coil is turned by 90.0 degrees to where the field line are perpendicular to its loops and the voltmeter registers a maximum voltage of 46.2 volts. Calculate the turning time.
Φ2 - Φ1 = NBAcos(0) - NBAcos(90) = NBA[ cos(0) - cos(90)] = NBA( 1 -0 ) = NBA
ΔΦm = (125)(2.17T)(0.00854m2) = 2.32 Tm2 ; E = ΔΦm / Δt ; Δt = ΔΦm / E
Δt = 2.32 Tm2 / 46.2 volts = 0.0502 seconds
When the N-pole of a bar magnet is pushed toward a coil, it induces a current in the coil. The current in the coil magnetizes the coil. The direction of the induced current is such that the end of the coil that is being approached by an N-pole becomes an N-pole itself to oppose the approaching N-pole. If the N-pole is moved away from the coil, the coil magnetizes again. This time the end of it that is near the receding N-pole, becomes a S-pole to oppose the going away of the bar magnet's N-pole. In a similar way:
When the S-pole of a bar magnet is pushed toward a coil, it induces a current in the coil. The current in the coil magnetizes the coil. The direction of the induced current is such that the end of the coil that is being approached by a S-pole becomes a S-pole itself to oppose the approaching S-pole of the bar magnet. If the N-pole is moved away from the coil, the coil magnetizes again. This time the end of it that is near the receding N-pole, becomes a S-pole to oppose the going away of the bar magnet's N-pole. The following figures indicate the polarization of the coil in different cases:
Lenz's law In brief:
The direction of the induced current in a coil is such that it opposes the motion of the external magnet that causes it.
Test Yourself 1:
Visualize you are in a class inside a huge horseshoe magnet that makes the ceiling the N-pole and the floor the S-pole of it. Also visualize a wire that carries electrons from right to left is in front of you. Answer the following questions:
1) The motion of electrons is (a) from left to right. (b) from right to left. click here.
2) As a result, the direction of motion of positive charges in the wire is (a) from left to right. (b) from right to left.
3) The force of the magnetic field on that current carrying wire inside it (a) pulls the wire toward you. (b) pushes the wire toward the board. (c) makes it move upward.
4) If the direction of the current in the wire reverses, it will then be (a) pushed toward you. (b) pushed toward the board. (c) pushed downward. click here.
Now, visualize that a big rectangular coil of copper wire (6ft long and 4ft high) is hanging from the ceiling and positive current is flowing through it clockwise.
5) The direction of the positive current in the upper side of the rectangle is (a) from left to right. (b) from right to left.
6) The top side of the coil will be pushed (a) toward you. (b) toward the board. click here.
7) The bottom side of the coil will be pushed (a) toward you. (b) toward the board.
8) The coil as a whole has therefore a tendency for (a) moving toward you. (b) moving toward the board. (c) rotation and this is the basis for making electric motors. click here.
Now, suppose you disconnect the battery that is feeding the coil and the current flowing through the coil drops to zero. Also, suppose the rectangular coil is attached to a horizontal shaft that passes through the midpoints of its 4-ft sides. If the shaft is turned quickly, some 30, 40 , or 50 degrees such that the top side of the coil comes toward you while the bottom side moves toward the board, answer the following questions: click here.
9) The direction of the positive current in the lower side of the coil will be (a) from left to right. (b) from right to left.
10) The direction of the positive current in the upper side of the coil will be (a) from left to right. (b) from right to left.
11) As a result, within that brief rotation of some 40 or 50 degrees, the direction of the current in the coil will be (a) clockwise. (b) counterclockwise.
12) The magnetic flux, Φm of a field of strength B through a surface of area A depends on (a) B and A only. (b) A only, (c) B and A as well as the orientation angle, θ. click here.
13) The formula for magnetic flux is (a) Φm = BAθ. (b) Φm = BAsinθ. (c) Φm = BAcosθ.
14) θ is the angle that (a) field lines make with surface. (b) field lines make with the vertical direction. (c) field lines make with normal to the surface. click here.
15) The normal line to a flat surface is (a) necessarily perpendicular to all lines that lie in that surface. (b) perpendicular to only one line of that surface. (c) perpendicular to only two lines of that surface.
Problem: Suppose it is raining with no wind blowing. The rain lines are vertical. Also suppose you are holding a picture frame horizontally under the vertical rain. Answer the following questions. click here.
16) The normal to the picture frame surface is (a) vertical. (b) horizontal. (c) oblique.
17) The angle that the normal to the surface (frame) makes with the rain lines is (a) 90º. (b) 0º. (c) 180º. (d) b & c are possibilities.
18) The rain vector direction is (a) upward. (b) downward. (c) neither a nor b. click here.
19) The angle that the rain vector makes (a) with normal to the bottom surface of the frame is 0º. (b) with normal to the top surface of the frame is 180º. (c) both a & b are correct. Note that normal to the bottom surface is downward and normal to the top surface is upward.
20) The value of cos(0º) is (a) 1. (b) 0. (c) -1. click here.
Problem: Suppose that the rain strength is B = 1600 drops/m2. Let the picture frame be (0.80m by .50m). This makes a surface area of A = 0.40m2. For the following cases, calculate the flux of rain through the picture frame.
21) Rain is perpendicular to the frames surface going out of its bottom surface. Angle θ is (a) 0º. (b) 90º. (c) 145º.
22) Using Φrain = BAcosθ, the flux through the frame is (a) 320drops. (b) 640drops. (c) 0.
23) Tilting the frame by 30º with respect to the horizontal, makes normal to the frame, n tilt by 30º with respect to the rain vector. θ becomes (a) 30º. (b) 60º. (c) 120º. click here.
24) The new rain flux through the frame becomes (a) 320drops. (b) 640drops. (c) 550drops.
25) Tilting the frame by 60º with respect to the horizontal, makes normal to the frame, n tilt by 60º with respect to the rain vector. θ becomes (a) 30º. (b) 60º. (c) 120º. click here.
26) The new rain flux through the frame (for θ = 60º) becomes (a) 320drops. (b) 640drops. (c) 550 drops.
27) Tilting the frame by 90º with respect to the horizontal, makes normal to the frame, n tilt by 90º with respect to the rain vector. θ becomes (a) 30º. (b) 60º. (c) 0º.
28) The new rain flux through the frame (for θ = 90º) becomes (a) 320drops. (b) 0. (c) 550 drops. click here.
29) Faraday's law is mathematically expressed as (a) E =- ΔΦm. (b) E =-ΔΦm Δt. (c) E =-ΔΦm /Δt
30) For electricity generation by an induction coil, there has to be (a) a change in magnetic flux through a coil or loop of wire. (b) just a flux through a coil even if it is not changing. (c) both a & b. click here.
31) The greater the flux change (ΔΦm), (a) the greater the induced voltage (E ). (b) the weaker the induced voltage (E ).
32) The faster the change (smaller Δt ), (a) the greater the induced voltage. (b) the weaker the induced voltage.
Problem: A coil of copper wire has 333 turns and each loop of it has an area of 1.25m2. It is placed inside a 0.245T uniform magnetic field such that field lines are perpendicular to the surface of its loops. The magnetic field is turned off and its strength goes to zero in 0.0422 seconds. Answer the following questions: click here.
33) While the magnetic field is on, the flux through the coil is (a) 0.306 Tm2. (b) 102 Tm2. (c) 0.0 Tm2.
34) While the magnetic field is on, the induced voltage in the coil is (a)102 Tm2/s. (b)102 volts. (c) 0.0 (d) a & b.
35) When the field is turned off, the change in Φm is (a) 102 Tm2. (b) -102 Tm2. (c) 0.
36) The induced voltage during this change of flux is (a) 24200 volts. (b) 242 volts. (c) 2420 volts. click here.
37) If the ohmic resistance of the coil is 48.4Ω, the current in the coil is (a) 5.00A (b) 50.0A (c) 25.0 A
38) The induced current is (a) large enough to cause a short electric shock. (b) not large enough to cause electric shock.
Example 5: You are facing a loop of wire in front of you in which the current can be either clockwise or counterclockwise. (a) You hold the S-pole of a bar magnet in your hand and move it toward the loop on its N-pole side. Will the direction of the induced current in the loop be cw or ccw? (b) When you pull the magnet back, what will the direction of the induced be? (c) If you hold the N-pole of the magnet and approach the loop by its S-pole what current direction do you expect? (d) when you pull your hand back what is direction of the induced current?
Solution: First answer all parts and then to check your answer click here.
An inductor is a coil of wire. We have already learned that any coil of wire that a current flows through it has magnetic properties. An inductor resists the change in the current that flows through it. As soon as a coil is connected to a battery, the current in the circuit changes from zero to a certain amount. The change in current causes a change in the magnetic flux. As soon as the change in magnetic flux occurs, a voltage develops across the inductor other than the battery voltage. The developed voltage has an opposite polarity compared to the battery voltage (Lenz's law). The faster the connection time or the shorter the change in current applied to an inductor, the greater the opposing voltage that develops across it. If we show the change in the applied current per unit of time as ΔI /Δt , and the opposing voltage that the inductor develops as Vinductor , we may write the proportionality of the two as
Vinductor = L (ΔI / Δt)
where L is the proportionality constant that depends on the physical characteristics of the inductor itself. L actually depends on the # of turns per meter of the inductor, n, the length of it , l , the cross-sectional area of its loops, A, and μo, the permeability of free space or approximately air. The formula is
L = μon2 A l
L is called the self-inductance or simply the inductance of the inductor and has units of Ω-sec. called "Henry."
As soon as the current in an inductor stabilizes and does not change any more, the opposing voltage drops to zero and the inductor behaves as if it does not exist. This is simply because of the fact that the ohmic resistance of an inductor is normally small and it does not cause any significant voltage drop in a circuit in which the current is constant. That is why inductors are treated as short circuits for non-varying currents.
In brief, the opposing voltage that an inductor develops is proportional to L and the time rate of change of the current in the inductor ΔI /Δt.
Example 6: An 8.0-cm inductor has 16000 turns and a diameter of 4.0cm. Its resistance is 2.4Ω. Find (a) its self-inductance, L. It is connected to a 12-V battery and the connection time is known to be 0.0040 seconds. Determine (b) its final current, and (c) the opposing voltage it develops during connection.
Solution: (a) L = μon2 A l ; L = (4π x 10-7 Tm/A)[(16000/0.080m)2]π(0.020m)2( 0.080m) = 5.05H
(b) I = V / R ; I = 12 volts / 2.4Ω = 5.0Amps
(c) Vinductor = L (ΔI / Δt) ; Vinductor = (5.05Ω-sec)[(5.0 - 0)A / 0.0040sec)] = 6300 volts
Example 7: A 250-mH inductor is in series with a 5.0-Ω resistor, a key, and a 10.0-V battery. The key is turned on and during the current jump from 0 to its stabilized amount, an opposing voltage of 43.0V is measured. Find the connection time. See the figure shown below.
When current stabilizes, the inductor acts as if it does not exists. This is because of the fact that the ohmic resistance of an inductor is very small and it therefore does not contribute to any voltage drop. Therefore the current the circuit reaches to will be I = V/R = 10.0V/5.0Ω = 2.0Amps.
Using V = L(ΔI / Δt) and soling for Δt yields:
Δt = L(ΔI / V) = (0.250H)(2 - 0)A / 43.0V = 0.012 sec.
When a rectangular loop of wire is spun in a uniform magnetic field, the current in the loop moves back and forth in a push-pull manner. The result is called the "alternating current" because the direction of the current keeps alternating with time. If the angular speed, ω , is constant, the way current or voltage changes with time follows a sinusoidal function. The following figures show the variations:
Note that this graph shows how the generated voltage in the loop changes with angle θ: V = Vmax sin( θ ).
But θ = ω t. V becomes:
V = Vmax sin( ω t).
This gives voltage as a function of time.
Suppose the loop starts turning counterclockwise (ccw) from a horizontal position (θ =0). At horizontal position the magnetic flux through the loop is maximum. As it turns ccw to the vertical position (θ = 90˚) , the flux through the loop approaches zero. Past 90˚, the flux keeps increasing to its maximum until the loop is horizontal again (θ =180˚). Past 180˚, the loop's orientation reverses causing the current changes direction. The result is an alternating current that changes direction every half period with an equation of the form V = Vmax sin(ωt).
Example 8: The equation of the alternating voltage we get at house electric outlets is V = 170sin(377t) where V is expressed in volts and t in seconds. Determine the (a) max. voltage and angular frequency, (b) frequency, (c) period, and (d) the max. current if a 10.Ω electric iron is in use.
Solution: (a) Comparing V = 170sin(377t) with the general form V = Vmax sin(ωt), we get: Vmax =170volts and ω=377rd/s.
(b) ω = 2πf ; f = ω / 2π ; f = 60.0Hz
(c) T = 1 / f ; T = (1 / 60) sec. ; (d) Vmax = R Imax ; Imax = 17. Amps
Faraday's Law and V = Vmax sin(ωt):
It is easy to show that equation V = Vmax sin(ωt) is the result of Faraday's law of electromagnetic induction.
Faraday's law is E = -ΔΦm / Δt where Φm = NBAcos( θ ). In Φm , the product NBA is constant for an electric generator in which a rectangular loop of area A with N turns rotates in a uniform magnetic field of intensity B. Only angle θ is not constant and changes with time. Since θ = ωt, we may write Φm = NBA cos( ωt ). We can show that the variations of cos( ωt ) with respect to time t can be written as
Δcos( ωt ) / Δt . Calculus proves that this is equivalent to - ω sin( ωt ). Substituting this in the E equation results in:
E = NBAω sin( ωt ). The factor NBAω is constant and is called E max . Therefore
E = E max sin( ωt ) is the equation for alternating voltage. E max is the same thing as Vmax and E the same thing as V.
Example 9: The rotor winding of an AC-generator is rectangular (42cm by 75cm) and contains 400. turns and rotates in a uniform magnetic field of 0.25T at 3600rpm. Calculate (a) the maximum voltage it generates and (b) write the equation of the alternating voltage for it.
Solution: ω = 3600 (rev / min) (min / 60sec)( 2π / rev) = 377 rd / s
(a) E max = NBAω ; E max = (400.)(0.25T)(0.42mX0.75m)(377 rd /s) = 12,000 volts
(b) E = E max sin( ωt ) ; E = 12,000 sin (377t)
Average Voltage, Average Current, and Average Power:
Voltage and current of an AC-source vary according to a sine function in each cycle. We may calculate mean values for each in each cycle. As you know, a sine function is positive in half a cycle and negative in the other half. The mean value in each cycle is mathematically zero. Zero does not reflect the actual value. One method of calculating the mean values of actual voltage, current, and power is to first square them, then find their mean values, and finally take the square root of those mean values. This method is called "root mean squaring" and the values so found are called the " root mean squares."
Look at the graphs of current, voltage, and power in 2 full cycles. Although the voltage and current are both positive in half cycle and both negative in the next half cycle, their product, the power, remains positive as shown. Using calculus, it is easy to show that the mean power is (1/2) of the maximum power.
Prms = (1/2)Pmax
Since P = VI, we may write the above equation as
Vrms Irms = (0.707)(0.707) Imax Vmax
This may be broken into two products as:
Vrms = 0.707 Vmax and Irms = 0.707 Imax .
Example 10: A 100.-watt light bulb is connected to a a 120.-V AC-source. Determine (a), (b), and (c) the rms values of power, voltage, and current, respectively. (d), (e), and (f), find the corresponding maximum values, respectively. (g) What is the resistance of the light bulb when hot.
Solution: (a) The 100. watt is itself an rms value. Electric devices working with AC currents are mentioned with their rsm power and not with the max. power. Aslo the 120-V AC is itself an rms value as well. Therefore, the answers to Parts (a) and (b) are:
(a) 100. watts ; (b) 120.volts ; (c) Irms = Prms / Vrms = 100w / 120v = 0.833 Amps.
(d) Pmax = 2Prms = 200.w ; (e) Vmax = Vrms / 0.707 = 170v ; ( f ) Imax = Irms / 0.707 = 1.18amps
(g) Vmax = RImax ; R = 170v / 1.18amp = 144 Ω.
A transformer is an electric device that is used to increase or decrease the voltage of an AC-source. It is made of two coils that are very closely packed to where one can pick almost the entire magnetic flux generated by the other. One coil is called the primary side and the other the secondary side. When an input voltage is given to the primary side, an output voltage develops at the secondary side. The input voltage must vary with time. This makes the magnetic field developed at the primary coil vary with time. The change in the field causes a change in flux. The change in flux is sensed by the secondary coil. As a result a voltage develops at the terminals of the secondary coil. If the number of loops of the secondary coil is more than that of the primary coil, it results in a voltage increase at the secondary side and the transformer is called a "step-up" transformer. In general the voltage ratio of secondary to primary is equal to the turns ratio of secondary to primary. See Figure below:
Actual transformers lose a portion of the input power in the forms of heat and electric flux leak. For an actual transformer,
Pout < Pin and therefore VsIs < VpIp.
For actual transformers the voltage ratio still equals the turns ratio; however, the reciprocal of the currents ratio does not equal the turns ratio.
Example 10: An ideal transformer has Np = 200. and Ns = 3400. turns. The primary is connected to a 7.00-V AC-source. It pulls 1.70 amps from the source. Find the output voltage and current.
Solution: Since the transformer is ideal Vs / Vp = Ns / Np or Vp / 7.00 = 3400. / 200. or Vs = 119 volts.
The currents ratio is Is / Ip = Np / Ns or Is /1.70 = 200. / 3400. or Is = 0.100 Amp.
Example 11: An actual transformer has Np = 1200. and Ns = 100. turns. The primary is connected to a 120.-V AC-source. It pulls 2.50 amps from the source. Find the (a) output voltage (b) the output current if it were ideal, and (c) the efficiency of the transformer if the output current is 27.0amps.
Solution: (a) Vs / Vp = Ns / Np or Vp / 120. = 100. / 1200. or Vs = 10.0 volts.
(b) Is / Ip = Np / Ns or Is /2.5 = 1200. / 100. or Is = 30.0 Amps.
(c) The transformer is expected to deliver (Is)ideal = 30.0 Amps and it actually delivers (Is)actual = 27.0Amps. It is therefore 27 / 30 = 90.0% efficient. We could also calculate the input and out put powers and then find the ratio as shown below:
eff. = Pout / Pin ; eff. = [(27.0A)(10.0V)] / [(2.50A)(120.V)] = 90.0%
Test Yourself 2:
1) An inductor is (a) a coil of wire. (b) a closed loop of wire. (c) an open loop of wire. (d) a & b. click here.
2) An inductor stores (a) electric energy. (b) magnetic energy. (c) both a & b.
3) An inductor has usually a small ohmic resistance. It acts as if it does not exist when (a) the current that flow through it is varying. (b) the current through it is constant. (c) the voltage across it is constant. (d) b & c. click here.
4) When the current through an inductor is constant, the magnetic field it generates is (a) constant. (b) varying. (c) 0.
5) When the current through an inductor is constant, the constant magnetic field it generates passes through its own loops. As a result the magnetic flux that flows through its loops is (a) constant (b) varying. (c) 0. click here.
6) According to Faraday's Law of Magnetic Induction, if magnetic flux through a coil is constant, (a) an emf is developed by the coil. (b) no emf is developed by the coil. (c) no induced voltage in developed in the coil by that flux. (d) b & c.
7) When the current through an inductor changes, the magnetic flux caused by that varying current (a) changes as well. (b) does not change. (c) becomes steady.
8) The change in the magnetic flux that flows through the loops of a coil (a) generates an induced voltages across the terminals of the coil. (b) makes the current in the coil constant. (c) both a & b. click here.
9) The faster the change in the current through a coil, (a) the greater the change in the magnetic flux through that coil. (b) the greater the induced voltage that develops in the coil. (c) both a & b.
10) By faster means (a) the greater the Δt. (b) the smaller the Δt. (c) the shorter the time. (d) both b & c.
11) The self-inductance of a coil is proportional to (a) the square of the number of turns per meter of it, n2. (b) the area of each loop, A. (c) the length of it, l.. (d) a, b, & c. click here.
12) The self-inductance of an inductor is (a) L = n A l.. (b) L = n2 A l.. (c) L = μon2 A l.
13) In magnetism, μo is (a) the permittivity of vacuum for electric field effect transmission. (b) the permeability of vacuum for magnetic field effect transmission. (c) the coefficient of friction in vacuum. click here.
14) The induced voltage in an inductor (a coil) is therefore proportional to (a) ΔI. (b) 1/Δt. (c) ΔI /Δt. (d) L, the self-inductance of the inductor. (e) c & d.
15) The induced voltage in an inductor as a result of a current change through it may be calculated by (a) VL=ΔI /Δt. (b) VL = L (ΔI / Δt). (c) VL = L (ΔI Δt).
16) The value of μo is (a) 4 x10-7 Tm/Amp. (b) 4π x10-7 Tm. (c) 4π x10-7 Tm/Amp. click here.
Problem: A coil (an inductor) is connected to a 12-V car battery via a 48-Ω resistor in series with it. Note that an inductor has a small ohmic resistance by itself. Connecting it to a car battery, melts it down (low or no resistance means great current). It must be put in series with an appropriate resistor to control the current through it. The circuit is then disconnected and the disconnection time is 0.00031s. Let L = 0.465Henry. Answer the following questions:
17) Neglecting the ohmic resistance of the coil, the current in the circuit is (a) 4.0A. (b) 0.25A. (c) 6.0A.
18) before disconnection the change in the current ( ΔI ), is (a) 4.0A. (b) 0.25A. (c) 0. click here.
19) During disconnection, the ratio ΔI /Δt is (a) 392 Amps/sec. (b) 806Amps/sec. (c) neither a nor b.
20) The induced voltage, VL = L (ΔI /Δt) , during disconnection is (a) 375 volts. (b) 475 volts. (c) 0.
21) When a coil spins in a uniform magnetic field, the induced current in it (a) is always in one direction. (b) alternates back and forth (in half a turn it flows in one direction, and in the next half turn, in the opposite direction). (c) flows in the direction of the coil spin. click here.
22) The equation of the alternating voltage induced in a spinning coil in a uniform magnetic field is of the form (a) a sinusoidal function of time. (b) E =Emaxsin(ωt). (c) a & b.
23) In equation E =Emaxsin(ωt), the value of Emax is (a) NBAω. (b) NBA. (c) NBAθ.
24) In equation E =Emaxsin(ωt), ω is (a) constant. (b) variable. (c) often 0. click here.
Problem: A 125-turn coil of surface area 0.20m2 rotates at 3600.rpm in a 0.0742-T uniform magnetic field. Answer the following questions:
25) The maximum induced voltage (Vmax or Emax ) across its terminal is (a) 6700volts. (b) 700.volts. (c) 3700 volts.
26) The equation of the induced alternating voltage is (a) E =[700.volts]sin(377t). (b) E =[6700.volts]sin(3600t). (c) E =(1/2)(3600)t2 + 125t !!!. click here.
27) For an alternating source, the voltage is (a) at times negative. (b) at times positive. (c) at times zero. (d) a, b, & c.
28) By voltage being "at times negative," it is meant that (a) the direction of current changes. (b) the polarity at the generator's terminals changes. (c) a & b.
29) For an alternating source, the current is (a) at times negative. (b) at times positive. (c) at times zero. (d) a, b, & c.
30) When an alternating source is connected to a resistor, the voltage and current are in phase relative to each other. By "in phase", it is meant that (a) they reach their maxima together. (b) they reach their minima together. (c) they become zero together. (d) a, b, & c. click here.
31) If the frequency of an alternating voltage is 10/s, it means that in each second (a) the polarity changes 10 times. (b) the polarity changes 20 times. (c) the current is in one direction 10 times. (d) the current is in the opposite direction 10 times. (e) b, c, & d.
32) In alternating current sources, power is always positive because (a) when voltage is positive, current is also positive. (b) when voltage is negative, current is also negative. (c) power is the product of voltage and current, and because of (a) and (b), the product is always positive.
33) Power in AC sources fluctuates between (a) a positive and negative amount. (b) two negative amounts. (c) zero and a positive maximum amount. click here.
34) If we simply average the voltage or the current of an AC source over each full cycle, the mean value becomes (a) zero. (b) positive. (c) negative.
35) The root mean square (rms) value of voltage or current for an AC source is (a) less then its max. value. (b) greater than its max. value. (c) equal to its max. value. click here.
36) It is not correct to think of the maximum of a varying quantity (a) to be greater than its mean value. (b) to be less than its mean value. (c) to be equal to its mean value. "Maximum of a varying quantity is always greater than its mean value."
37) For an AC source, Vrms is (a) 1.414Vmax. (b) 0.5Vmax. (c) 0.707Vmax.
38) For an AC source, Imax is (a) 1.414Irms. (b) 0.5Irms. (c) 0.707Irms. click here.
39) For an AC source, Prms is (a) 1.414Pmax. (b) 0.5Pmax. (c) 0.707Pmax.
40) The AC voltage of about 120V that we may measure at a wall electric outlet is (a) Vmax. (b) Vrms. (c) Vavg.
41) The 75-watt power written on a light bulb, for example, is it (a) Pmax. (b) Prms. (c) Pavg. click here.
42) When a 100-watt light bulb is in use, in each voltage cycle, (a) there is only one instant that the power is 200 watts. (b) there are two instances at which the power is 200 watts. (c) there are two instances at which the power is zero. (d) b & c.
43) When a light bulb is on at a wall electric outlet, (a) at any instant, we may think of the rms voltage to be 120V. (b) there is an instant in each cycle at which the voltage is 170V. (c) there is an instant in each cycle at which the voltage is -170V. (d) there are two instances at which the voltage is zero. (e) a, b, c, & d.
44) A transformer can amplify (a) an AC voltage. (b) a DC voltage. (c) both a & b. click here.
45) Transformers are made such that the voltage ratio of secondary to primary is (a) always equal to the turns ratio of secondary to primary. (b) always equal to the turns ratio of primary to secondary. (c) neither a nor b.
46) Transformers are made such that the current ratio of secondary to primary is (a) always equal to the turns ratio of secondary to primary. (b) always equal to the turns ratio of primary to secondary. (c) neither a nor b. click here.
47) If a transformer is assumed to be an ideal one, the current ratio of secondary to primary is (a) may be set equal to the turns ratio of secondary to primary. (b) may be set equal to the turns ratio of primary to secondary. (c) neither a nor b.
48) The efficiency of a transformer is defined as (a) the turns ratio of secondary to primary. (b) the input-to-output power ratio. (c) the output-to-input power ratio. click here.
49) The efficiency of an ideal transformer is (a) 1 (b) 0. (c) 2.
50) Efficiency greater than 1 (a) is not possible, because it violates the law of conservation of momentum. (b) is not possible, because it violates the law of conservation of energy. (c) is possible for ideal transformers because the output voltage can be greater than the input voltage. (d) is possible for ideal transformers because the output current can be greater than the input current. click here.
Problem: For a transformer, Vp = 120V, Ip = 0.025A, and Vs = 8.0V. Answer the following questions.
51) The input power is (a) 120 watts. (b) 3.0 watts. (c) 48 watts. click here.
52) If the transformer were an ideal one, its output power would be (a) 3.8 watts. (b) 3.0 watts. (c) 4.2 watts.
53) If the transformer were an ideal one, its output current would be (a) 0.6 Amps. (b) 1.0 Amp. (c) 0.375 Amps.
54) If the actual output current is 0.36 Amps instead, the actual output power is (a) 2.88 w. (b) 3.5 w. (c) 3.4 w.
55) The efficiency of this transformer is (a) 0.92. (b) 0.96. (c) 0.68. click here.