Chapter 21
The Giancoli onLine Tutor

Electromagnetic Induction and Faradays Law; AC Circuits

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This chapter like the previous one and the one that follows it deals with material best covered at a mathematical level higher than what is available here. As in the previous chapter, the key to solving the problems in this chapter is finding the right equation to use. Keep in mind, though, that the key to choosing the right equation is understanding some complex relationships.

I. Magnetic Induction

In the last chapter we learned that electric currents produce magnetic fields and that changing electric currents produce changing magnetic fields . The converse is also true (i.e., changing magnetic fields produce electric currents), and forms the conceptual basis of this chapter.
Changing magnetic fields induce an emf (which in its turn can produce a current).

When magnetic field lines (only one of many is shown in the figure) cross an area, A, the magnetic flux, F, is defined as the perpendicular component of the field multiplied by the area. In most problems the magnetic field (B) will be perpendicular to the area, in which case
F = BA.
The unit of flux is the weber (Wb), which equals 1Tm². The magnetic field may also be given in units of Wb/m². The induced emf (

) is given by

= -DF/Dt.
The minus sign indicates that the induced emf opposes the change that produces it. This is called Lenz's Law (read about it in Giancoli). When calculating an induced emf, ignore the minus sign and decide later on the proper direction.

Changing Flux in a Loop

Most problems involve a current loop in which the flux is changing because either the area of the loop changes or the magnetic field changes (or maybe both). In problems like this you will usually be given an initial and final value of A or B or F and a time Dt. Calculate the change in F and divide by Dt. If the number of turns in the coil (N) is given, multiply DF/Dt by N. You may be asked to find the current in the loop given the resistance or the resistance given the current. In either of these cases use Ohm's Law.

Interactive Example
Given a 0.05 m² 5-turn loop placed in a magnetic field whose maximum value is 0.5 T, how fast must the magnetic field change ( DB/Dt in tesla/s) in order to induce an emf of 12 V across the loop?
Enter your answer then click outside the box.

Moving Conductor in a Magnetic Field

A variation on the loop problem above is shown at the left. A metal bar slides along a U-shaped metal track. The blue region contains a magnetic field and the flux through the loop defined by the rectangle a b c d is increasing because the size of the loop is changing owing to the rods motion. The induced emf is given by
= Blv,
l is the length of the rod and v is the velocity of the rod. If you are asked to find the resistance or current, use Ohm's Law, as before. Actually, the metal track does not need to be there because any metal object (such as an airplane wing) will have an induced emf across it as it goes through a magnetic field.

Interactive Example
How fast must a 2-meter long metal rod move through a 0.5 T magnetic field in order to induce an emf of 12 V across the length of the rod?
Enter your answer then click outside the box.

Electric Generators

Electric generators are built around a wire coil with N turns and area (A) rotating with a frequency (f; usually 60 Hz) in a magnetic field (B). This produces an alternating voltage given by
= coswt, where w is its angular frequency, t is the time and

is the maximum (or peak) emf. Angular frequency is given by = 2pf (and is usually 370 s^-1), and

given by
= NBAw.
Although the equations seem complicated, the problems are actually easier than the problems in earlier chapters. Most of the problems give four of the elements

(peak emf), N (number of turns), B (magnetic field), A (loop area) or w (angular frequency) and ask you to find the other using the above equation.

Interactive Example
How many turns must a 0.05 m² coil have in a 0.5 T magnetic field in order to induce an emf of 120 V a frequency of 60 Hz ?
Enter your answer then click outside the box.

Transformers

An ideal transformer is a 'square' metal ring with windings connected to it on two sides. The number of turns in the windings is N_p on the primary side of the transformer and N_s on the secondary side. An alternating voltage (V_p) applied to the primary side produces an alternating voltage (V_s) in a secondary circuit. The relationship between the voltages and the number of turns in the windings is given by V_s/V_p = N_s/N_p.
If the transformer is 100% efficient then the power into the transformer equals the power out of the transformer and since power equals voltage (V) times current (I), V_sI_s = V_pI_p for the transformer and I_s/I_p = N_p/N_s.
A real transformer is not 100% efficient, however, so its efficiency, e, is given by
e = V_sI_s/V_pI_p
These equations and Ohm's Law can be used to eliminate the secondary voltage and current to give a very useful relationship between the primary current, the primary voltage, and the resistance. If R is the resistance in the secondary circuit, then an effective resistance (R_eff) can be defined as R_eff = e(Np/Ns)²R . Based on these relationships,
V_p =I_pR_eff.
Most problems involving transformers can be solved using one of the equations presented here.

Interactive Examples
A transformer has 50 turns in the primary and 250 turns in the secondary. If the transformer is 75% efficient and the secondary is connected to a resistance of 20 ohms, what is the effective resistance of the primary?
Enter your answer then click outside the box.

If the primary voltage is 120V, what is the current in the primary circuit?
Enter your answer then click outside the box.

Variations on this problem type may involve power transmission. Power (P), which is given VI, is transmitted from one point to another over wires. In real life the wires have a resistance (R) and so they experience a heat loss given by RI². The ratio of heat loss to power transmitted is
RI²/VI = RI/V = RP/V².
Because R and P are fixed, the greater the transmission voltage the lower the power loss. For example, if the transmission voltage (V) is increased by a factor of 10, then the power loss (P) decreases by a factor of 100.

II. Inductance

A transformer works because the changing current in the primary coils creates a changing magnetic field in proportion to the number of turns in the primary coil. Additionally the changing magnetic field 'trapped' in the metal core creates a changing (alternating) emf in the secondary coil which is proportional to the number of turns in the secondary coil. This is an example of mutual inductance. Mutual inductance occurs when an emf,

, is induced in a second coil by a changing current in another coil. The equation is
= - MDI₁/Dt
where M is a proportionality constant, I₁ is the current in the other coil and t is the time. The proportionality constant (M) is called the mutual inductance. The unit for M is called the henry (H) and equals 1Ws .

More important to solving problems in this section is the concept of self inductance, L. Self inductance, which also has units of henry, is the magnetic interaction of a coil with itself. A coil, now called an inductor and symbolized by

resists an increase or decrease in current by inducing an opposing emf. In this way it acts much like a resistor with the important exception that it does not dissipate energy. The defining equation for L is
= - LDI/Dt
, where

is emf, I is current, and t is time. For a simple solenoid -long coil of length, l, number of turns, N, and cross sectional area, A,-the self inductance (L) is given by
L = m_oN²A/l ,
where m_o, the permeability of free space, is equal to 4px10^-7 Tm/A.

Interactive Example
An inductor with a diameter of 2 cm, a length of 10 cm, and 200 turns has an inductance to the nearest mH of...?
Enter your answer then click outside the box.

III. LR Circuits

In a simple direct current LR circuit (shown at right):
I = (V/R)(1 - e^-t/t),
where I is the current, V is the voltage, R is resistance, t is the time, and t = L/R is the time constant, the time it takes the current to reach 63% of its final value I_max = V/R. L is the self inductance.

Interactive Activity
In the pictured circuit, R = 10 W and the selected inductance (L) is 30 H, so the time constant (t) is 3 seconds and the maximum current (I_max) is 10 A. Stop the simulation at 3 seconds and verify the above numbers. Pick another value for L, predict the time constant and check your result by running the simulation. To run the simulation, close the switch in the circuit .

t= seconds I= amps

L =

A similar problem involves a circuit with an established current flowing and an inductance. When the switch in the circuit is opened, the current will not instantly drop to zero. It will instead drop to 37% of its initial value in 1 time constant. Experiment by changing the value of L, then opening the switch, and seeing how the current changes.

t= seconds I= amps

L =

In a AC circuit the inductor acts as virtual resistance called the reactance (X_L). The reactance is given by X_L = 2pfL, where f is the frequency of the source (normally 60 Hz) and L is the inductance. And the analog to Ohm's Law (V=IR) is
V=IX_L
If the current (I) in such a circuit is
I = I_ocos2pft,
then the voltage is given by
V = -V_osin2pft,
where V_o=I_oX_L. Because (cos2pft)(sin2pft) averages over one cycle to zero, no power is expended.
Both V_o and I_o are the peak values for the voltage and current, respectively. In AC circuits, it is the rms values which are normally used. You may want to review the relevant material in Chapter 18.

Interactive Example
If the rms voltage is 120 volts at 60 Hz and L = 0.318, what is the rms current? Enter your answer then click outside the box.

IV. AC Circuits

A similar situation occurs in a AC capacitor circuit. The capacitive reactance is given by X_C = 1/(2pfC) where the frequency is f and C is the capacitance. If the current is again given by
I = I_ocos2pft,
the voltage is given by
V = V_osin2pft.
Notice that this is similar to, but different from the voltage for the inductor. In a capacitor circuit, the current is said to lead the voltage by 90^o. In an inductor circuit, the current is said to lag the voltage by 90^o.

Interactive Example
If the rms voltage is 120 volts at 60 Hz and C =1 mF what is the rms current? Enter your answer then click outside the box.

V. LRC Series Circuits

The diagram shown on the right depicts an LCR circuit and has many interesting properties. An LCR circuit has an impedance (Z) which is a combination of resistance (R) and reactance (X), and is given as follows
Z=√(R²+(X_L-X_C)²).
"Ohm's Law" for an LCR circuit is
V_rms=I_rmsZ or V_o=I_oZ.
To be specific in presenting the following information we will assume that

V_rms= 100 V

V_o= 141 V

L= 20 mH

C= 20 mF

R= 20 W

f= 400 Hz

Using the equations above we find that

X_L = 50.27 W
X_C = 19.9 W
Z = 36.36 W
I_rms = 2.75 A
I_o= 3.9 A
The rms voltage is given by
V_Lrms = I_rmsX_L = (2.75 A)(50.27 W) = 138.25 V
V_Crms = I_rmsX_C = (2.75 A)(19.9 W) = 54.7 V
V_Rrms = I_rmsR = (2.75 A)(20 W) = 55 V
The rms voltages on each element don't add to the rms applied voltage of 100 V because the instantaneous voltages are out of phase with each other. If the current is alternating as a cosine function, the instantaneous voltage across the whole circuit will be given by
V=V_ocos(2pft+f) where f is the phase angle and is given by f = tan^-1((X_L-X_C)/R) = tan^-1(30.37/20) = 56.6^o

Interactive Activity
The simulation below steps through one complete cycle (t=0 -> t=.0025= 1/400) in 25 steps and displays instantaneous voltages and the current, as well as the instantaneous power (P = VI). The average power is given by
P_ave= I_rmsV_rmscos(f)= 151.25 W
The accumulating average power is the last entry. Over 1 cycle it will approximate the above value and it would approach it exactly if smaller steps where used . Notice that...

the overall instantaneous voltage (V) is always the sum of the instantaneous voltages across the three elements,

the voltage across the resistor is always in phase with the current,

sometimes the instantaneous power is negative but on the average is positive.

t=
V_L=
V_C=
V_R=
V=
I=
P=
P_ave=

One use of a LCR circuit is as a resonant or tuning circuit. As the frequency is changed the I_rms current equal to V_rms/Z will increase to V_rms/R when X_L = X_C and consequently Z = R. This will occur when the frequency (f ) is equal to f_o given by
f_o= (2p)^-1√(1/LC)
where L is the inductance and C is the capacitance.

If L =30 mH and C=15 mF then f_o= 218 Hz. The smaller the value of R the more dramatic the height and narrowness of the peak. This allows you to select a particular frequency. The Q of the circuit is a measure of the ability of the circuit to tune a particular frequency. It is defines as f/Df, where f is the frequency and Df is the frequency width at half maximum as shown in the figure at the right. Try it below.

Interactive Activity
You can select different values for R and the frequency will be scanned from 138 to 300 Hz in steps of 4 Hz.

R= ohms frequency: Hz I= amps

What is the Q of the circuit when R = 10 W? Find Df by finding the difference in frequency between the two frequencies at which the current is at half its maximum. Enter your answer below then click outside the box.
Try the same experiment with R = 1 W. You will find it much more difficult than with R = 10 W. The answer for Q in this case should be 30.