PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
PHYS 155: Final Tutorial
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
Eric Peach
Saskatoon Engineering Students’ Society
eric.peach@usask.ca
BREAK
Faraday’s Law
of Induction
Inductance
Summary
April 13, 2015
Overview
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
1 Magnetic Fields
2 The Three Right Hand Rules
3 Force and Torque On Wire
BREAK
Force and
Torque On
Wire
4 Ampere’s Law
Ampere’s Law
BREAK
5 Faraday’s Law of Induction
Faraday’s Law
of Induction
Inductance
6 Inductance
Summary
7 Summary
Tutorial Slides
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
These slides have been posted:
sess.usask.ca
homepage.usask.ca/esp991/
Section 1
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Magnetic Fields
Magnetic Fields
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
Magnetic Fields have no start, no finish. There is NO point of origin.
∇·B=0
They are created by movement of charge.
BREAK
Force and
Torque On
Wire
Ampere’s Law
∇ × B = µ 0 J + µ 0 E0
Changing fields can incite EMF in other objects.
BREAK
Faraday’s Law
of Induction
Inductance
Summary
∇×E=−
dB
dt
dE
dt
Magnetic Field Lines
PHYS 155:
FINAL
Magnetic Field lines have no start, no finish.
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Magnetic field strength B related to density of field lines.
Three Right Hand Rules
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
Very Spatial
Assume Positive Charge and Current
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Find magnetic field direction in various cases
Find force on a moving charge or current carrying wire.
RHR 1: Magnetic Field Around a Wire
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Magnetic field circles around wires, in a direction CCW when wire is coming
out of page.
Put your thumb along wire.
Your fingers curl around wire along magnetic field lines.
RHR 2: Magnetic Field Inside a loop or Coil of Wire
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Wire wound in a coil generates a magnetic field directed along the axis of the
coil.
Fingers wrap around the loop or coil, in the direction of the current.
Thumb points in the direction of magnetic field.
RHR 2: Additional Consideration
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
We can find RHR No. 2 by applying Rule No. 1 to a coil instead of a straight wire!
RHR 3: Force on a Moving Charge
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
Charge or current moving in a magnetic field experiences a magnetic force.
BREAK
Perpendicular to direction of movement
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Perpendicular to magnetic field.
RHR 3: Force on a Moving Charge
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Thumb Along Direction of Movement
Index Finger along Magnetic Field
Palm / middle finger points to the force!
Remember, this is for POSITIVE Charge. For negative charge like
electrons, the force is opposite!
RHR 3: Force on a Moving Charge
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Right Hand Rules: Summary
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
RHR #1: Straight Wire
Thumb Along Direction of Current
Fingers curl the way the magnetic field does.
RHR #2: Coil / Loop of Wire
Curl fingers around coil of wire in direction of current
Thumb tells you direction of magnetic field
RHR #3: Charge Moving in Magnetic Field
Thumb Along Movement of Positive Charge
Faraday’s Law
of Induction
Index finger along magnetic field direction
Inductance
Palm or middle finger tells you direction of force on positive charge.
Summary
Some Helpful Magnetic Formulas
PHYS 155:
FINAL
Magnetic Field for a long, straight wire
Eric Peach
Magnetic
Fields
|B| =
Magnetic Field at Centre of Loop
The Three
Right Hand
Rules
|B| = N
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
µ0 I
2πr
µ0 I
2R
Magnetic Field Inside Solenoid
N
I
`
Force on Moving Charge / Wire in Magnetic Field
|B| = µo nI = µ0
Inductance
F = q(v × B)
Summary
F = qvB sin θ
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Example 1 – Lorentz Force
A proton moving at 4.00 × 106 m/s through a magnetic field of magnitude 1.70 T
experiences a magnetic force of magnitude 8.20 × 10−13 N. What is the angle
between the proton’s velocity and the field?
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
Example 1 – Solution
Draw a picture showing this effect. Lorentz force is
The Three
Right Hand
Rules
~ = q~v × B
~
F
BREAK
Force and
Torque On
Wire
so
~ | = qvB sin θ
|F
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Using magnitude of force, velocity and magnetic field, solve for θ.
Answer: θ = 48.83◦ or 131.2◦ .
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
Example 2 – Cyclotron Motion
The Three
Right Hand
Rules
A singly charged ion of mass m is accelerated from rest by a potential difference
∆V . It is then deflected by a uniform magnetic field (perpendicular to the ion’s
velocity) into a semicircle of radius R. Now a doubly charged ion of mass m0 is
accelerated through the same potential difference and deflected by the same
magnetic field into a semicircle of radius R 0 = 2R. What is the ratio of the masses
of the ions?
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Example 2 – Solution
Write the kinetic energy of the particle.
1
KE = q∆V = mv 2
2
Write the conditition for centripetal acceleration
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
(1)
Fm = Fc ,
qvB =
mv 2
R
(2)
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
Example 2 – Solution
Sub (1) into (2) and solve for m in terms of q, B, ∆V and R.
The Three
Right Hand
Rules
m=
BREAK
Force and
Torque On
Wire
B2
qR 2
2∆V
Solve for m0 in terms of q, B, ∆V and R. Note that all you have to do is
tweak your q, B & R values.
Ampere’s Law
BREAK
m=
Faraday’s Law
of Induction
Inductance
Summary
Answer: m0 = 8.00 × m.
B2
(2q)(2R)2
2∆V
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Example 3 – Hold Straight Condition
A velocity selector consists of electric and magnetic fields described by the
~ = B ĵ, with B = 15.0mT. Find the value of E such that
expressions ~E = E k̂ and B
a 750-eV electron moving in the negative x direction is undeflected.
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
Example 3 – Solution
Draw your picture. Which way will the magnetic field try to direct the
electron? (Up)
Which way does the electric force need to apply? (Down). Which way should
the field be pointed? (Up)
Solve for electron velocity using KE = 12 mv 2
Draw FBD and write the condition for no deflection:
BREAK
Faraday’s Law
of Induction
Inductance
Summary
F~M + F~E = 0
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
Example 3 – Solution
Figure out the magnitude of the magnetic force on the electron.
~ = 3.90 × 10−14 k̂ N
F~M = q~v × B
BREAK
Force and
Torque On
Wire
Note that
F~E = E k̂ = −F~M
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Solve for E.
Answer: 244 kV/m
Break
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
See you in 10 Minutes
Section 4
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Force and Torque on Wire
due to Magnetic Field
Force on a Wire
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Extension of RHR #3.
~
F = I (~` × B)
F = I `B sin θ
` is length of wire, whose direction is the direction of the current.
For 2 parallel wires, the force of attraction/repulsion is
|F| =
µ0 I1 I2
L
2π d
Faraday’s Law
of Induction
Inductance
Summary
Attraction if current in same direction
Repulsion if current in opposite direction
Torque on a Loop
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
A loop of wire in a magnetic field expriences a torque, according to
~ × B]
~
~τ = NI [A
τ = NIAB sin θ
the A vector has magnitude equal to area of loop, direction points in direction
of magnetic field generated by loop (use RHR 2).
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Example 4 – Torque on Wire
A current of 17.0mA is maintained in a single circular loop of 2.00 m
circumference. A magnetic field of 0.800 T is directed parallel to the plane of the
loop. What is the torque exerted on the loop by the magnetic field?
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
Example 4 – Solution
1
Draw a picture of the loop of wire and the field. Conceptualize: if current
flows in the loop, and the magnetic field is parallel to the plane of the loop,
which side will try to lift up? Which side will be pushed down?
2
Solve for area of the loop using circumference.
3
Use Torque Formula to solve for torque.
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
~ × B]
~
~τ = NI [A
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Answer: 4.33 ×10−3 N·m.
PHYS 155:
FINAL
Example 5 – Attraction Between Wires
Eric Peach
The current in the long, straight wire is I1 = 5.00 A and the wire lies in the plane
of the rectangular loop, which carries a current I2 = 10.0 A. The dimensions in the
figure are c =0.100 m, a = 0.150 m, and ` = 0.450 m. Find the magnitude and
direction of the net force exerted on the loop by the magnetic field created by the
wire.
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
PHYS 155:
FINAL
Example 5 – Solution
Eric Peach
Magnetic
Fields
1
Ask yourself: Which way will each part of the loop try to move?
2
Find the strength of the magnetic field due to the long wire at the left side of
the loop.
µ0 I1
B=
2πc
Find the Lorentz force on the left side of the loop
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
3
~ = I2 ~` × B
~
F
Ampere’s Law
BREAK
Faraday’s Law
of Induction
4
Find the magnetic field strength and the force on the right side of the loop.
Inductance
Summary
B=
µ0 I1
2π(c + a)
~ = I2 ~` × B
~
F
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
Example 5 – Solution
BREAK
Note that the top and bottom parts of the loop experience forces too, but
they cancel out.
Force and
Torque On
Wire
Subtract the force of the right side of the loop from the force of the left side
of the loop to find net force.
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Answer: −2.7 × 10−5 î N
Section 5
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Ampere’s Law
Ampere’s Law
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Let’s us find the magnetic field around an enclosed loop of wire. Comes from
∇ × B = µ 0 J + µ 0 E0
dE
dt
H
~ · d`
~ = µ0 Ienc .
This simplifies to B
Pick some sort of symmetry, usually circular, around some source of current. Then
you can find the magnetic field strength.
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Example 6 – Ampere’s Law
A packed bundle of 100 long, straight, insulated wires forms a cylinder of radius R
= 0.500 cm. If each wire carries 2.00 A, what are the (a) magnitude and (b)
direction of the magnetic force per unit length acting on a wire located 0.200 cm
from the centre of the bundle?
PHYS 155:
FINAL
Example 6 – Solution
Eric Peach
1
Recall Ampere’s Law:
I
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
2
3
Force and
Torque On
Wire
~ · d`
~ = µ0 Ienc
B
Figure out the current enclosed by a radius of 0.2cm. (It’s 32 amps).
~ will be
If we integrate around a circle, centred at the centre of the wire, B
constant because of radial symmetry. So you get
B · (2πr ) = µ0 Ienc
Ampere’s Law
BREAK
Faraday’s Law
of Induction
4
~
Solve for B
Inductance
5
Multiply by 2A, the current of 1 wire at that radius. (F/`=IB)
Summary
Answer: F/` = 6.4 mN.
Break
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
See you in 10 Minutes
Section 7
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Faraday’s Law of Induction
and Lenz’s Law
Lenz’s Law
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
A loop or coil of wire that is exposed to a
change in magnetic flux through the loop will
induce an EMF whose corresponding magnetic
field will oppose the original change in flux.
Faraday’s Law of Induction
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Mathematical Representation of Lenz’s Law
E = −N
dΦ
d
= −N (AB cos θ)
dt
dt
So we get Back EMF for one of three reasons: Change in Area, Magnetic Field
Strength or Angle.
Usually only one will change at a time in any given question.
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Example 7 – Faraday Law of Induction
A coil formed by wrapping 50 turns of wire in the shape of a square is positioned in
a magnetic field so that the normal to the plane of the coil makes an angle of 30.0◦
with the direction of the field. When the magnetic field is increased uniformly from
200 µT to 600 µT in 0.400 s, an emf of magnitude 80.0 mV is induced in the coil.
What is the total length of the wire in the coil?
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
Example 7 – Solution
1
Recall Faraday Law of Induction
The Three
Right Hand
Rules
E = −N
dΦ
d
= −N (AB cos θ)
dt
dt
BREAK
Force and
Torque On
Wire
2
Here, only the magnetic field is changing, so this simplifies to
E = −NA
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
3
dB
cos θ
dt
Solve for A, then find side length, then multiply by 4N to get length of wire.
Answer: 272m.
Bar and Rail Problems
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
1
Figure out the induced EMF
Lenz’s Law to find direction.
Faraday’s Law for magnitude.
The Three
Right Hand
Rules
BREAK
2
Using EMF, find current in loop (Ohm’s Law, KVL)
Force and
Torque On
Wire
3
Using I and Lorentz force law, find magnetic force on bar.
4
Draw FBD of Bar, solve unknowns. Find any input force required.
5
Watch out for directions of B, v and F, especially on angled rails!
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
PHYS 155:
FINAL
Example 8 – Bar and Rail
Eric Peach
The picture shows a bar of mass m = 0.200 kg that can slide without friction on a pair of
rails separated by a distance ` = 1.20 m and located on an inclined plane that makes an
angle θ = 25.0◦ with respect to the horizontal. The resistance of the resistor is R = 1.00
Ω and a uniform magnetic field of magnitude B = 0.500 T is directed downward,
perpendicular to the ground, over the entire region through which the bar moves. With
what constant speed v does the bar slide along the rails?
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
PHYS 155:
FINAL
Example 8 – Solution
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
1
Solve for EMF using E = −`vB cos θ
2
Use Ohm’s Law to find current through loop. I = E/R
3
Lorenz Force Law (F = qvB) to find the back EMF force on the bar (* Watch
out for direction here *).
4
Draw FBD to find relation between FM , N and Fg .
5
For constant velocity, you can relate FM and Fg . Solve for v
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
v=
Faraday’s Law
of Induction
Inductance
Summary
Answer: 2.80 m/s
mg tan θR
`2 B 2 cos θ
Section 8
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Inductance
Inductance
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Self Inductance
E = −N
dΦ
dt
EL = −L
dI
dt
Mutual Inductance
Force and
Torque On
Wire
Es = −M
Ns Φs = MIP
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
L = µ0
Energy In an Inductor
1
U = LI 2
2
dIP
dt
N2
µ0 µr N 2 A
A=
`
`
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
Example 9 – Mutual Inductance
The Three
Right Hand
Rules
Two solenoids A and B, spaced close to each other and sharing the same cylindrical
axis, have 400 and 700 turns respectively. A current of 3.50 A in solenoid A
produces an average flux of 300 µWb through each turn of A and a flux of 90.0
µWb through each turn of B. (a) Calculate the mutual inductance of the two
solenoids. (b) What is the inductance of A? (c) What emf is induced in B when
the current in A changes at a rate of 0.5 A/s ?
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
PHYS 155:
FINAL
Example 9 – Solution
Eric Peach
Magnetic
Fields
1
NS ΦS = MIP
The Three
Right Hand
Rules
BREAK
Use the mutual inductance formula that has all the variables you know.
2
Force and
Torque On
Wire
To find self inductance, use self inductance formula relating flux and current.
NΦ = LI
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
3
To find EMF, use EMF formula with mutual inductance.
ES = −M
dIP
dt
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Example 9 – Solution
It might be helpful to pay attention to the directions of the coils here. Remember
that the ’-’ serves to remind you that it is back EMF, but the winding of the coil
actually decides the direction of current.
Answer: 0.018 H, 0.0343 H, -9mV.
Transformers
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Just remeber the transformer turn ratio formula, relating current, voltage and
number of turns in the Primary and Secondary coils.
VP
IS
NP
=
=
NS
VS
IP
Use Ohm’s Law and P = VI to solve for everything else.
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Example 10 – Transformers
In the transformer shown, the load resistance RL is 50.0 Ω. The turns ratio N1 /N2
is 2.50, and the rms source voltage is ∆Vs = 80.0 V. If a voltmeter across the load
resistance measures an rms voltage of 25.0 V, what is the source resistance Rs ?
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
Example 10 – Solution
Remember the big important transformer equation.
The Three
Right Hand
Rules
VP
IS
NP
=
=
NS
VS
IP
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Using the secondary coil voltage and Ohm’s law, find current in secondary coil.
Using turns ratio, find the current in primary coil and voltage in primary across
the transformer.
KVL to find voltage across resistor, and Ohm’s Law to find resistance.
Answer: 87.5 Ω
Summary
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
1
Go through the slides (both mine and the ones on blackboard) and make up a
solid formula sheet.
2
Have a good section on unit conversions, and what units you can expect from
certain formulas.
3
Do the assignments. Each question has a recipe to solve it. Practise your Lenz
Law to anticipate direction of magnetic fields and back EMF.
4
Watch out for weird units on the exam.
BREAK
Force and
Torque On
Wire
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary
Good Luck!
PHYS 155:
FINAL
Eric Peach
Magnetic
Fields
The Three
Right Hand
Rules
These slides have been posted:
BREAK
sess.usask.ca
Force and
Torque On
Wire
homepage.usask.ca/esp991/
Ampere’s Law
BREAK
Faraday’s Law
of Induction
Inductance
Summary