Chapter 31 and 32

Magnetic Induction:

When a charge moving at 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 that is placed inside the field of a horseshoe magnet, as shown below:

Fig. 1

In the left figure, key K is open, no current flows in the loop; therefore, there is no force on the wire segment in the magnetic field.  In the right figure, key K is closed, current I flows in the loop (charges 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 I develops in the loop.  This can be verified by placing a sensitive ammeter in the loop as shown below:

Fig. 2

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 develops in the loop (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.

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 use 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.

Fig. 3

Magnetic field lines 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 loop of 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:     

Fig. 4

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 = NBA 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 30.0o angle with the loop surface as shown.  The magnetic field has a strength of 1.2T.  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 30.0o as shown.   The θ we want is the angle that B makes with the normal to the surface.  

θ = 90.0o - 30.0o = 60.0o and the flux is

Φm = BAcosθ =1.2T (0.001m2)cos(60.0o) = 6.0x10-4 Tm2.

 or,   Φm = 6.0x10-4 Weber.

Fig. 5

 

Example 2:  A student experimenting with magnets turns a circular loop of wire, 4.0cm in radius, from a horizontal position to a vertical position in between a 0.17T horseshoe magnet.  Calculate the change in the 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)( cos0 ) = 8.5x10-4 Tm2.

 When the loop is vertical, θ = 90o, and the flux is

 Φ2 = B A cosθ = (.17 T)(0.0050m2)( cos90 ) = 0.

 Therefore, the change in magnetic flux is:

 Φ2 - Φ1 =  0 - 8.5x10-4 Tm2  = - 8.5x10-4 Tm2.

            ΔΦ = - 8.5x10-4 Tm2.

Fig. 6

 

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:

Fig. 7

To generate electric current in a coil, the magnetic flux through its loops must change.  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 following formula:

In Faraday's formula, V is the induced voltage or electromotive force (emf) generated in 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 V has the unit of volt as is expected.  The (-) sign indicates that the direction of the induced current opposes the direction of change in the flux that causes itThis 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 1.36T occurs.  Find the absolute value of the induced voltage across the coil.  The magnetic field lines stay perpendicular to the loops during the process.

Solution: ΔΦm = NBAcosθ     ;    ΔΦm = (750)(1.36 T)(π)(0.060m)2 cos(0)  =  11.5 Tm2.

        |V|= ΔΦm / Δt  =  11.5 Tm2 / 0.50s = 23 volts.

 

Example 4:  A 125-loop coil of wire is in a 2.17T 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.

 Solution:  Φ2 - Φ1  NBAcos(0) - NBAcos(90)  =    NBA[ cos(0o) - cos(90o)]  =  NBA( 1 -0 ) = NBA.

 ΔΦm = (125)(2.17T)(0.00854m2) = 2.32 Tm2     ;    |V| = ΔΦm / Δt    ;    Δt = ΔΦm / V 

Δt =  2.32 Tm2 /46.2 volts  =  0.0502 seconds.

 

Lenz's Law:

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 approach of that N-pole.   If the N-pole is moved away from the coil, the coil gets magnetized 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 approach of that S-pole.   If the S-pole is moved away from the coil, the coil gets magnetized again.  This time the end of it that is near the receding S-pole, becomes a N-pole to oppose the going away of the bar magnet's S-pole.  The following figures indicate the polarization of the coil in different cases:

Fig. 8

Lenz's law: 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 sitting in a classroom facing the board.  Suppose the class ceiling is the N-pole and floor the S-pole of a huge horseshoe magnet.  Also visualize a wire in front of you that carries electrons from right to left.  Answer the Questions 1 through 4.

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 positive current 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 metal wire (6ft long and 4ft high) is hanging from the ceiling and positive current is flowing through it clockwise.  Answer Questions 5 through 8.

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 Questions 9, 10, and 11  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 this 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 vertically.  Also suppose you are holding a recangular frame horizontally under it.  Answer the following questions:   click here.

16) The normal to the frame surface is (a) vertical  (b) horizontal  (c) oblique.

17) The angle that the normal to the surface of the frame (n) makes with the rain lines is (a) 90o   (b) 0o   (c) 180o   (d) b & c are two options.

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 0    (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 (intensity) is B = 1600 drops/m2/sec.   Let the 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 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/sec     (b) 640drops/sec     (c) 0.

23) Tilting the frame by 30o,  makes normal to the frame, n, also 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/sec  (b) 640drops/sec   (c) 550drops/sec.

25) Tilting the frame by 60o,  makes normal to the frame, n, also 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/sec     (b) 640drops/sec     (c) 550 drops/sec.

27) Turning the frame by 90o,  makes normal to the frame, n, also tilt by 90o with respect to the rain vector.  θ becomes  (a) 30o     (b) 60o     (c) 90o.

28) The new rain flux through the frame (for θ = 90o) becomes (a) 320drops/sec     (b) 0     (c) 550 drops/sec.   click here.

29) Faraday's law is mathematically expressed as (a) V =- ΔΦm    (b) V =-ΔΦm Δt    (c) V =-ΔΦm /Δt.

30) To generate current in a coil, there must be (a) a magnetic flux change in the coil's loops   (b) just a flux through the 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 (V)  (b) the weaker the induced voltage (V).

32) The faster the change (smaller Δt ),  (a) the greater the induced voltage   (b) the weaker the induced voltage.

 Problem: A coil has 333 loops and each loop has an area of 1.25m2.  It is placed inside a 0.245T uniform magnetic field such that field lines are perpendicular to the planes 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    (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) 25.0A.

38) The induced current is (a) large enough  (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.

 

Inductors:

An inductor is a coil of wire.  We have already learned that any current carrying coil has magnetic properties.  In fact, an inductor is a device that stores magnetic energy. 

Inductors Resist or Respond to Current Changes:

Any time the current in an inductor is set to change, the inductor resists that change by developing an opposing voltage.   As soon as a coil is connected to a battery, the current in the circuit changes from 0 to I as much as say, ΔI.  The current change ΔI causes a flux change ΔΦm in the coil.   The flux change ΔΦm causes a voltage to develop across the coil other than the battery voltage.  The developed voltage has an opposite polarity compared to the battery voltage (Lenz's law).  Experiment shows that the shorter the connection time Δt,  the greater the developed opposing voltage across the coil.  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 VL , we may write the proportionality of the two as

VL = L (ΔI /Δt)

where L called the "self-inductance" of the inductor is the proportionality constant.   L depends on the physical characteristics of the inductor.   L actually depends on the number of turns per meter of the inductor n,  its length l,  its loop cross-sectional area A,  and μo.   The formula for self-inductance L is

 

L = μon2 A l

 The SI unit of L is  Ω-sec called "Henry."

 μo = 4π x 10-7 Tm/A is of course, the permeability of vacuum.

Fig. 9

The opposing voltage developed by the battery does not last longer than the connection time.  The connection time is a very small fraction of a second.  As soon as  the current in the inductor stabilizes and does not change with time, the opposing voltage drops to zero and the inductor behaves as if it does not exist.  This is simply because inductors usually have small ohmic resistances and do not cause significant voltage drops in circuits.   For this reason, inductors can be used as short circuits for non-varying currentsIn brief, the opposing voltage that an inductor develops is proportional to L and the time rate of change of current, ΔI /Δt.

Example 6:  An 8.0cm long inductor has 1600 turns and a diameter of 4.0cm.  Its resistance is 2..  Find (a) its self-inductance, L.   It is connected to a 12V 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:  Note that n is the number of loops per meter or simply the N/l of the inductor.  Here n = 1600/0.080m.

(a)  L = μon2 A l  ;  L = (4π x 10-7 Tm/A)[(1600/0.080m)]2 π(0.020m)2( 0.080m) = 0.051H.

 (b)  I = V / R   ;    I = 12 volts / 2.4Ω = 5.0A.

 (c)  VL = L (ΔI / Δt)  ;  VL = (0.051Ω-sec)[(5.0 - 0)A / 0.0040sec)] = 64V.

 

Example 7:  A 250mH 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.

 Solution: 

 When current stabilizes, the inductor acts as if it does not exists because of its negligible ohmic resistance.  The only resistance in the circuit will therefore be only the 5.0Ω resistor when current stabilizes.  We may calculate this current as   I = V/R = 10.0V/5.0Ω = 2.0Amps.  This means that the current changes from 0 to 2.0A.

 Using VL = L(ΔI/Δt)  and sovling for Δt yields:

 Δt = L(ΔI/V)  = (0.250H)(2 - 0)A/43.0V = 0.012 sec.  This means that the opposing voltage of 43.0V does not last more than 0.012s.  It just quickly goes to that peak and then gone!

 

Fig. 10

 

Alternating Current:

When a rectangular loop of wire is spun in a uniform magnetic field, the developed 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 or reversing with time.  At a constant angular speed ω, the way current in the spinning coil changes with time is a sinusoidal function.  It means that its mathematical equation is either a sine or a cosine function of time.  The following figures show the variations:

Fig. 11

Explanation:

Suppose the loop starts turning CCW from a vertical position with its normal to the loop, n, pointing horizontally to the left.    At the start of rotation, the angle that n makes with B is θ = 90o.  At this initial vertical position, the magnetic flux through the loop is zero (cos 90o = 0).   As the loop turns CCW to the horizontal position, the normal to the loop (vector n) becomes vertically downward where its angle with B will be (θ = 0o), and the flux through the loop becomes maximum (cos0o = 1).   Past 90o, the horizontal loop rotates to become vertical.  During this 2nd quarter rotation, flux decreases and becomes zero when the loop is vertical again with θ =180o.  Past 180o, the loop's orientation reverses causing current I to change direction.  Past 180o,  normal to the loop n starts becoming upward and opposite to B resulting in negative flux.  This makes the loop current change direction resulting in an alternating current that changes direction every half turn with an equation of the form: 

V = Vmax sin(ωt). 

 

     The graph on the right shows how the generated voltage across the loop changes with angle θ:   V = Vmax sinθ.

 But since  θ = ωt, the Voltage equation becomes:

    V = Vmax sinωt.  This equation gives the loop's voltage variation as a function of time.

 

Fig. 12

 

 

The Derivation of V = Vmax sin(ωt):

The voltage generation is the result of the flux change m through the loops of a coil during time interval dt.  Applying Faraday's law 

{V = -dΦm /dt }  to  [ Φm = NBAcosθ ]   with  θ = ωt,  we get:

V = -dΦm /dt = - d[NBA cos(ωt)] /dt   =    NBAω sin(ωt).  or,

V = NBAω sin(ωt).  

We write this as  V = Vmax sinθ     where    Vmax= NBAω.

 Note that  d/dt{cos(ωt)} = -ωsin(ωt)The factor NBAω is a constant and is nothing but the maximum voltage,  Vmax"

Question:  In V = NBAω sin(ωt), what would you come up with if you try to find the maximum value of V?  Of course, you know that N, B, A, and ω are constants of a certain design generator.   The only variable is sin(ωt).  How do you determine the maximum value of V if NBAω is constant?

Example 8:  The equation of the alternating voltage we get at city electric outlets is V = 170sin(377t) where V is expressed in volts and t in seconds.   Determine the (a) maximum voltage, (b) angular frequency, (c) frequency, (d) period, and (e) maximum current if a 12Ω electric iron is in use.

Solution:  (a) Comparing  V = 170sin(377t) with the general form V = Vmax sin(ωt), we get:  

(a)  Vmax = 170volts  ;    (b)  ω=377rd/s.

(c) ω = 2πf    ;    f = ω / 2π     ;    f =  60.0Hz.

(d) T = 1/f    ;    T = (1/60)s    ;     (e)  Vmax = R Imax     ;    Imax  = 170V/12Ω  = 14A.

 

Example 9:  The rotor winding of an AC-generator is a rectangular coil (42cm by 75cm) that contains 400 loops and rotates in a uniform magnetic field of 0.25T at 3600rpm.  Calculate (a) the Vmax of the generator and (b) write the equation of its alternating voltage.

Solution: ω = 3600 (rev/min)  =  3600 (6.28 rd)/(60s)  =  377 rd /s.

(a) Vmax = NBAω    ;    Vmax =  (400)(0.25T)(0.42mX0.75m)(377 rd /s) = 12,000 volts

(b) V = Vmax sin( ωt )     ;    V = 12,000 sin (377t)   where V is in volts and t in seconds.

 

Average Voltage, Average Current, and Average Power:

Both voltage and current in AC sources are sinusoidal.  We may calculate a mean value for each within one cycle.  As you know, a sine function is positive in 1/2 cycle and negative in the next half cycle.  The mean value in each cycle is mathematically zero.  Mathematically, the negative half completely cancels the positive half.  Of course, ZERO does not reflect the actual value.  We know that the overall voltage, current, and power can't be zero.  One method of calculating the actual mean value for voltage, for example, is to first square its values in each segment, then find the mean value of the squares, and finally take the square root of that mean value.  This way, by squaring, there won't be any negatives to result in a zero mean value. This method is called "root mean squaring" and such mean value is called the " root mean square (rms)."

On the right, look at the graphs of current, voltage, and power each shown for 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.         (1)

The proof will follow after using this Prms to come up with relations for  Vrms  and  Irms.

 The factor 1/2 in  Eq. (1), may be written as:

 

Fig. 13

Since P = VI, we may write Eq. (1) as

Vrms Irms = (0.707)(0.707) Vmax Imax 

This may be broken into two products as:

Vrms  = 0.707 Vmax          ;          Irms = 0.707 Imax.        (2)

Example 10:  A 100 watt light bulb is connected to a 120V AC-source.  Determine (a), (b), and (c) the rms values of power, voltage, and current, respectively.   (d), (e), and (f), find their corresponding maximum values, respectively.   (g)  Calculate the in-use resistance of the light bulb.

Solution:  (a) The 100 watt itself is an rms value.   Electric devices designed for AC currents are labeled with their rsm power and not with the maximum power.   Also, the 120V AC itself is an rms value as wellThis means that the answers to Parts (a) and (b) are already given.

(a) Prms = 100w    ;    (b) Vrms =120V    ;    (c)  Irms  =  Prms / Vrms  =  100w/120V = 0.833A.

(d) Pmax = 2Prms = 200w    ;    (e) Vmax = Vrms / 0.707 =  170V    ;    ( f )  Imax = Irms / 0.707 = 1.18A.

(g) Vmax = RImax    ;    R = 170v / 1.18amp = 144 Ω.

 

Transformers:

A transformer is an electric device that is used to increase or decrease the voltage of varying sources (mainly, AC-sources).  It is made of two coils that are very closely packed and share the same iron core.  Sharing the same iron core means sharing the same magnetic flux Φm as well.  One coil is called the primary side and the other, the secondary side.  When a varying voltage is given to the primary side, an output varying voltage develops at the secondary side.  The input voltage must vary with time.  This makes the magnetic field B developed at the primary coil vary with time, accordingly.  The changes in B cause changes in flux Φm.  The changes in Φm are sensed by the secondary coil.  As a result a varying voltage develops at the terminals of the secondary coil.  If the number of loops of the secondary coil Ns,  is more than that of the primary coil Np, it results in a voltage increase at the secondary side and the transformer is called a "step-up" transformer. 

In general, for all transformers, the voltage ratio of secondary to primary  Vs/Vp  is equal to the loops ratio of secondary to primary Ns/Np.  See Fig. 14.

Fig. 14

Actual transformers lose a portion of the input power in the forms of heat and magnetic flux leak.  For an actual transformer,

Pout < Pin             and therefore         VsIs < VpIp.

For actual transformers the voltage ratio still equals the loops ratio; however, the reciprocal of the currents ratio does not equal the loops ratio.

Example 10:  For a transformer that is assumed to be ideal,  Np = 200 and Ns = 3400 loops.  Its primary side is connected to a 7.00V AC-source and draws a current of 1.70A.  Find both the output voltage and current.

Solution:   Since the transformer is assumed ideal:   Vs/Vp  = Ns/Np    ;    Vp/7.00 = 3400/200    ;   Vs = 119V.

The currents ratio is:   Is/Ip = Np/Ns     ;      Is/1.70 = 200/3400      ;      Is = 0.100A.

 

Example 11: An actual transformer has Np = 1200 and Ns = 100 loops.  The primary is connected to a 120V AC-source.  It draws 2.50A 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    ;   Vp/120 = 100/1200    ;     Vs = 10.0V.

                (b)  Is/Ip = Np/Ns     ;     Is/2.5 = 1200/100     ;     Is = 30.0A.

(c) The transformer is expected to deliver  (Is)ideal = 30.0A and it actually delivers (Is)actual = 27.0A.   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)(120V)] = 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 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 flux that flows through its loops is (a) constant (b) varying   (c) 0.   click here.

6) According to Faraday's Law, if the magnetic flux through a coil is constant, (a) an emf develops in the coil   (b) no emf develops in the coil   (c) there is no induced voltage across the coil.  (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 through the loops of a coil (a)  induces a voltage across the coil  (b) induces a constant current in the coil   (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 across the coil   (c) both a & b.

10) By faster we mean (a) a greater Δt    (b) a smaller Δt    (c) a shorter 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) its length, 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 opposing voltage an inductor develops is proportional to (a) ΔI only   (b) 1/Δt only   (c) ΔI /Δt    (d) L, the self-inductance of the inductor   (e) c & d.

15) The opposing voltage an inductor develops 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/A    (b) 4π x10-7 Tm     (c) 4π x10-7 Tm/A.    click here.

Problem: A coil (an inductor) is connected to a 12V car battery via a 48-Ω resistor in series with it.  (Due to the small resistance of the inductor, this 48-Ω resistor is placed in series with it to limit the current draw from the battery).  The circuit is then disconnected and the disconnection time is 0.00031s.   Let L = 0.465Ω-sec.   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 ΔI /Δt  is (a) 392 A/s   (b) 806 A/s   (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   (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 (a)  a sinusoidal function of time    (b)  V=Vmaxsin(ωt)     (c) a & b.

23) In equation V=Vmaxsin(ωt), the value of Vmax is (a) NBAω    (b)  NBA     (c) NBAθ.

24) In equation V=Vmaxsin(ωt) ω is (a) constant     (b) variable    (c) often 0.    click here.

Problem: A 125 loop coil with a loop surface area of 0.20m2 rotates at 3600 rpm in a 0.0742T uniform magnetic field.  Answer the following questions:

25) The maximum induced voltage (Vmax across its terminal is (a) 6700volts     (b) 700.volts     (c) 3700 volts.

26) The equation of the induced alternating voltage is (a) V = [700 volts]sin(377t)    (b) V = [6700volts]sin(3600t)    (c) V = (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 AC 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 correct to think of the maximum of a varying quantity (a) to be less than its mean value   (b) to be greater than its mean value   (c) to be equal to 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 its   (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) For real transformers, Vs/Vp  is (a) always equal to  Ns/Np   (b) always equal to  Np/Ns   (c) neither a nor b.

46) For real transformers, Is/Ip  is (a) always equal to Ns/Np   (b) always equal to Np/Ns   (c) neither a nor b.    click here.

47) If a transformer is assumed to be an ideal one,  the current ratio Is/Ip  (a) may be set equal to the loops ratio Ns/Np  (b) may be set equal to the loops ratio of Np/Ns  (c) neither a nor b.

48) The efficiency of a transformer is defined as (a) the loops 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) An 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.     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) 0.6As   (b) 1.0A  (c) 0.375A.

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.