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Phy 122 Notes e'
Section 5: Magnetic Fields & Forces
There are two kinds of magnetic poles, called north and south. (N and S are to magnetism what + and - are to electricity.) Opposites attract, likes repel.
A compass is a little magnet on a pivot. It feels a torque (gets rotated) in the field of another magnet.

The magnetic field vector, B , points in the direction a compass would point.
(Field lines outside a magnet run N to S.)
Magnetic force on an electric charge:
A stationary charge feels no force:
Moving charge does feel a force, of magnitude
So, you could define B as units: 10
4
gauss = 1 tesla
F = q v B sinθ

Direction of this force (on a + charge): Perpendicular to both v rule:

and B , according to the right hand

Right hand rule: With v

and B sticking out of the origin, wrap fingers around
 origin pointing from v

toward B . Or, as at right.

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Examples ( means vector pointing toward you, x means it's pointing away from you.)
Problem 5-1: The beam of electrons in a CRT has a velocity of 1.6 x 10
7
m/s east. Earth's field here is B = 60 μT north. Find the force on an electron in the beam, including direction.
Ans: 1.54 x 10
-16
N, down (toward the ground)
Problem 5-2: In each case, state whether there is a force on the charged particle. If there is, say whether its direction is +x, -x, +y, -y, +z or -z.
(Applications: Electrons in a vacuum tube, such as the magnetron in a microwave oven. Ions in a mass spectrometer. Other…)
Charges moving through a wire, instead of empty space, feel this same force. So, there is a force on a current carrying wire.
F = I l
B sinθ

I = current, l
= length, B = magnetic field, θ = angle between wire and B .

Direction: Right hand rule, as above. Direction of v is direction (positive) current flows.
(Applications: This force makes motors go around. ...)
Problem 5-3: Is there a force on the wire? If there is, say whether its direction is +x, -x, +y, -y, +z or -z.
Problem 5-4: A power line carries 50 A in the +x direction. Earth's field (.6 gauss) points down the z axis. Find the magnitude and direction of magnetic force on a section of line 65 m long.
Ans: .195 N, +y

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Magnetic fields exist around electric currents. Special case:
Long straight wire:
Field lines around a long straight wire are circles.
(Magnetic field lines never have a beginning or end.)
Which way they go around is given by the right hand rule: Point thumb along the current, fingers wrap around like the field lines.
B = μ
0
I
2πr
μ
0
= permeability of free space
= 4π x 10
-7
T·m/A
Magnetic materials: All magnetic fields come from electric currents. In a permanent magnet, these currents are within its atoms: the movement and spin of the charged particles the atom is made of.
(In a non-magnetized substance, these subatomic magnets have random directions, so when you add them up over the whole object, their total is zero.)
Example 5-a: Find the force on wire D due to C, if each is 1.0 m long. Include its direction.
Problem 5-5: The motion of an electron and the current in a wire are both parallel to the x axis as shown. Find the force on the electron, including its direction.
Ans: 2.50 x 10
-17
N, +y

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Section 6:
Solenoid = a long coil:
Field outside a solenoid is quite weak. (B 0)
Field inside is uniform, and equal to
N = number of turns B = μ
0
IN
L
(In text: B = μ
0
In where n = N/L.)
Direction (any coil or a loop of wire): Grab coil with right hand, fingers wrapping around in direction current goes.
Thumb points in direction of field inside.
Also on formula sheet: B of a flat (i.e., short) coil.
Example 6-a: An electron inside the solenoid is going left at 2.3 x 10
7
m/s. Find the force on it, including its direction.
Problem 6-1: In A - C, state whether there is a force on the charged particle. If there is, say whether its direction is +x, -x, +y, -y, +z or -z.
Example: a generator.
Free charges in the wire feel a force (q v B sinθ), which makes them flow as a current.
Induction:
In D & E, is there a force on the straight wire? If there is, say whether its direction is
+x, -x, +y, -y, +z or -z.

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More convenient to think in terms of an induced voltage rather than force on electrons:
Faraday’s law of Induction:
Magnetic Flux: = B cosθ A
N = number of turns in coil.
“Amount of field” passing through the loop. unit: weber (1 T = 1 Wb/m
2
)
A = area of loop
θ = angle between and axis of the loop:
In a picture, is represented by the number of field lines passing through an area. B is how close together the field lines are.
Example 6-b: For the spinning loop of wire at the top of the page, θ changes from 40° to 50° in
.00046 s. The area of the loop is 5.0 cm
2
and the field is .150 tesla. Find the average induced voltage during this time.
Problem 6-2: The magnet moves toward the loop along its axis in a way that steadily increases B from 0 to 2.0 mT over 2.0 sec. What does the meter read during those two seconds?
Ans: 2.83 μV
Problem 6-3: If that loop has a resistance of 5.0 x 10
-4
Ω (without the meter), what is the current?
Ans: 5.65 mA
Generator: As the loop at the top of the last page spins, B and A stay constant, but θ changes. A voltage is induced, by Faraday’s law.
Transformer: A pair of coils arranged so that same magnetic flux goes through both. Changing current in "primary" coil creates a changing magnetic field which induces a voltage in "secondary" coil.

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The purpose of a transformer is to change voltages.
Ideally, there is no resistance, so the induced emf is the only voltage across each coil:
Problem 6-4: a) A cell phone charger is to be plugged into 120 V AC, and have an output of 5.00 V
AC. If its transformer has 1000 primary windings, how many turns should the secondary have?
Ans: 42 b) If it was plugged into 120 V DC instead, how many volts would its output be?
Self - Induction:
Faraday's law says an emf is induced in any loop of wire with a changing flux through it. This includes the loop creating that flux in the first place:
It can be shown from Faraday's law that L
I
E = t
I = current
L = coil’s self inductance (unit: henry.)
(A more rapidly changing current creates a more rapidly changing magnetic field, which induces a larger voltage. L depends on the construction of the coil.)
Problem 6-5: (a) The switch is left closed for long enough to reach a steady current. How much current? (b) The switch is then suddenly opened, stopping the current in .010 second. How large is the emf induced in the coil?
Ans: 10 A, 600 V (Enough to throw an arc across the switch, or even shock you.)

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Section 7:
LR circuits:
Current decay: At t = 0, the switch is quickly thrown from A to B.
Find I as a function of time.
The coil opposes the decrease in current, so it uses energy from its magnetic field to keep I flowing.
The current tapers off with time according to:
I = I
0 e
-t/τ where
τ = L/R
(e = 2.71828... Use e x
button.)
τ = L/R = time constant = time to get about
63% (1-1/e) of the way to the final value.
(When I ≈ 37% of I
0
in this case.)
Current building up: If you start with no current, then connect to the battery at t = 0, coil holds current back from final value:
I = (
E
/R)(1 - e
-t/τ
)
Example 7-a: The switch is left at A for a long time, then suddenly thrown to B. Find: (a) the time constant, (b) the current when the switch was still at A, (c) the current after one time constant, (d) the current after two time constants, and (e) the current 6.0 ms after the switch was closed.
Problem 7-1: A 13 mH coil has a resistance of 8.0 Ω. There is no current in it until, at t = 0, it is connected to a 12 V car battery. Find: (a) the time constant, (b) the final current, as the time gets very large, (c) the current after one time constant, (d) the current 6.0 ms after it began.
Ans: 1.625 ms, 1.50 A, .945 A, 1.46 A

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Discharging:
RC circuits behave similarly to LR circuits:
Just as the current in a coil decreases with time, so does the charge in a capacitor: q = Q
0 e
-t/τ
τ = time constant = RC
If you need the voltage across the capacitor:
V = q/C from the definition of capacitance.
(Make the substitution Q o
/C = V o
)
If you need the current, put that voltage into Ohm’s law.
Charging up: q = C
E
(1-e
-t/τ
)
Find V from C = q/V
Find I from loop rule and Ohm’s law
Problem 7-2: A 10 μF capacitor, originally charged to 500 V, is connected to a 1.0 MΩ resistor.
Find: ( a) the time constant, (b) the voltage after 35 s and (c) the current at this time.
Ans: 10.0 s, 15.1 V, 1.51 x 10
-5
A
Problem 7-3: An uncharged 10 μF capacitor is placed across a 9.0 V battery with an internal resistance of .50 Ω. Find the voltage across the capacitor 3.0 s after it was connected.
Ans: 4.06 V

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LC circuits:
1. Close switch. C starts discharging.
2. C is discharged. L's emf keeps I flowing.
3. L's emf reaches 0, so I reaches 0. C is re-charged.
4. And, so on.
Charges move back and forth like a mass bouncing on a spring.
Charge on capacitor:
q = Q max
cos(ωt + )
Find V from C = q/V
T = Period = Time for one complete vibration.
Frequency (f) = the number of vibrations per second.
1 cycle per second = 1 Hertz f = 1
T
1
LC
where
ω = 2 f
is the frequency in radians per second, f is the frequency in cycles per second. is called the "angular frequency", f is just the "frequency".
phase angle = 0 if clock starts when q = Q max
Problem 7-4: A 10 μF capacitor, originally charged to 500 V, is connected to a 7.50 mH coil at t =
0. Find: (a) the period and ( b) the voltage after 1.50 ms.
Ans: .00172 s, 346 V

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Section 8: Alternating Currents i = I max
cos (ω t) v = V max
cos (ω t + ) where ω = 2 f = phase angle
What does "120 V AC" mean? It's the effective voltage; the DC voltage that would deliver the same power.
(It's not the average voltage. That's zero.)
Since power = i
2
R = v
2
/R, the average values of i the peak values: (i
2
) av
= I max
2
2
and v
2
are what matter. These average out to half
/2 and (v
2
) av
= V max
2
/2. Take the square root: and
Average power loss in a resistor:
RMS = root mean square: Square then average then take square root.
AC meters read RMS values.
Problem 8-1: An AC voltmeter reads 121 V from a 60 Hz outlet. Write the equation for how v varies with time, if = 0.
Resistors:
Since v = iR, V
RMS
= I
RMS
R
For v = iR to be true at all times, v and i must rise and fall together. v and i are in phase.

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Capacitors:
In the DC case (sec. 7), current flows at first, but drops off as the capacitor gets charged.
(Charge in capacitor repels current.) The voltage across the capacitor builds up from zero.
With AC, things start the same way, but continue as cosine/ sine waves: v lags ¼ cycle
(90°) behind i.
1
V
RMS
= I
RMS
X
C
where X
C
Capacitive reactance (unit: )
C
(The greater the frequency the less ohms: At high f, charge flows in for less time before flowing out again, making average charge on capacitor less, making less repulsion. The greater the capacitance, the less ohms, because a bigger capacitor has room in it for more charge.)
Problem 8-2: Find the RMS current for a 6.0 μF capacitor in a household plug. (Assume "household plug" means 60 Hz, 120 V)
Ans: .271 A
Problem 8-3: Repeat with 600 Hz. (ans: 2.71 A)
Inductors (coils):
V
RMS
= I
RMS
X
L
where X
L
= ωL Inductive Reactance ( )
Current is opposed by the coil's self-induced emf. A bigger frequency means more ohms, because a more rapidly changing current induces more emf (
E
= -L t). A larger inductance also induces more opposition, by
E
= -L t .
Phase: Apply loop rule to this: v
C
+ v
L
= 0
v
C
= -v
L
So, if v
C
is a sin, as above, v
L
is a -sin (upsidedown sin):

- 28 - v leads i by 1/4 cycle (90 )
To help remembering this: Eli the ice man: ELI ICE
E is for voltage, I for current. L (inductor): E leads, I follows. C (capacitor): I leads, E follows.
RLC Series Circuit:
Effect of the phase differences:
If you add v
C
+ v
L
, they cancel because they’re out of phase. (If V
C
is 5 V and V
L
is 4 V, the
“total” of the two is 1 V not 9 V.) If you include v
R
, its phase is in between, so it does something between adding and subtracting:
Behaves like:
=
Phasor diagram phase angle for circuit as a whole

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Problem 8-4: If V
R
= 20 V, V
L
= 40 V and V
C
= 62.36 V, what is the voltage across the entire group?
Ans: 30.0 V
Divide that equation through by I; make some substitutions:
V/I = R
2
( X
L
X
C
)
2
Define Impedance: Z = R
2
( X
L
X
C
)
2 unit:
V = IZ
Example 8-a: Find (a) the RMS current in this circuit, and (b) the phase angle.
Problem 8-5: Find I and .
Ans: .197 A, -60.5°
Resonance:
Around a certain frequency, a large current can get through. (Other f s are stopped by L or C.)
Current is largest when Z is smallest.
R
2
( X
L
X
C
)
2
is smallest when X
L
= X
C
L = 1/ C
1
Resonant frequency where ω = 2 f
LC
Problem. 8-6: An RLC circuit is used as a radio's tuner. If L = 7.96 μH, what should C be to receive a station at 1000 kHz?
Ans: 3.18 nF

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Review of Sec. 5 – 8:
1. What is the voltage across the resistor 200 s after the switch is closed?
Ans: 4.18 V
2. An RMS current of 4.5 A flows through a 3.0 mH coil when it is connected to a 50 V (RMS), 400
Hz AC source. What is the coil's internal resistance?
Ans: 8.16 Ω
3. The coil seen here from the side has 700 turns and encloses an area of .0300 m
2
. What is the average voltage induced in this coil during this .930 ms?
Ans: 477 V
4. On a current balance, similar to the one you used in lab 5, there are 15 strands of wire on the base, and 9 strands hanging on the bottom of the balance. These nine strands are 17.0 cm long, and an average distance of 1.30 cm from the wires on the base. If the current through all the wires is
4.50 A, how large is the force on the bottom of the balance?
Ans: .00715 N
5. Short answer, 5 points each: a. Atoms of both iron and chromium are tiny magnets. Yet, a piece of iron can be strongly magnetized, while a piece of chromium can't. What can happen in iron but not chromium which causes this difference? b. A discharging capacitor in an RC circuit starts out with 10.0 V between its plates. What is the voltage after one time constant? c. You have a capacitor and an inductor, both with negligible internal resistance. If you connect either of them to a household outlet (120 V, 60 Hz), the 20 A fuse in the circuit will blow after a moment. If connected to 120 V DC instead, i. would the capacitor still blow the fuse? ii. would the inductor still blow the fuse? d. Is there a force on the straight wire? If there is, say whether its direction is +x, -x, +y, -y, +z or -z. e. A proton moving north through a magnetic field feels a force toward the east. What is the direction of the magnetic field?