NEXT MIDTERM EXAM
The second common hour exam will be held on Thursday NIGHT Nov 13, 9:5011:10 PM in 4 locations on 2 campuses. The assignment to exam room is based
on the first letters of your last name:
Allison Road Classroom 103 (Busch)
Physics Lecture Hall (Busch)
Lucy Stone Hall A102 Aud (Livingston)
Beck Hall 100 Aud (Livingston)
Aaa-Gzz
Haa-Lzz
Maa-Rnn
Roa-Zzz
There will be optional review sessions on Thursday, November 13; details will
be announced.
All exams will be no calculator, closed book exams with only 1 page of
equations. Types of questions include iclickers, only formulae, pure concepts,
simple numbers. For details go to
http://www.physics.rutgers.edu/ugrad/227/intro.html#Examinations.
If you have a conflict you have to contact Professor Cizewski
Cizewski@physics.rutgers.edu at your earliest opportunity but not later than
midnight on Sunday, November 9 to request a conflict exam.
1
Lecture 18. Motional EMF
Outline:
Motional EMF.
Magnet falling in Cu pipe revisited.
Eddy Current Brake.
3
Lenz’s “Law” (better Rule) one more time
𝑑
� 𝐸 ∙ 𝑑𝑙⃗ = −
𝑑𝑑
𝑙𝑜𝑜𝑜
�
𝑠𝑢𝑢𝑢𝑢𝑢𝑢
𝑑𝑙⃗
x
𝐵 ∙ 𝑑𝐴⃗
𝑑𝐴⃗
1. Choose the direction of circulation – e.g.,
clockwise.
2. Corresponding 𝑑𝐴⃗ - into the board (righthand rule).
3. The flux decreases → the induced field
has to be into the board.
4. Thus, in order to have ∮
𝐸 ∙ 𝑑𝑙⃗ > 0,
𝑙𝑜𝑜𝑜
𝐸 should be oriented along 𝑑𝑙⃗.
Lenz: the current induced in the loop produces the magnetic field that tries to
compensate (unsuccessfully) the change in the magnetic flux
At home, analyze the problem by choosing the direction of circulation counterclockwise.
8
Motional EMF
𝐼
𝐹𝐵
𝑣
𝐹𝑢𝑢
Primitive model of a DC generator.
Part of the loop is in a uniform B field.
The loop is pulled to the right.
Find the induced current I.
1. Faraday’s Law approach:
𝑑Φ 𝑑
ℇ =
=
𝐵𝐵𝐵 = 𝐵𝐵𝐵
𝑑𝑑
𝑑𝑑
Current:
𝐼=
ℇ 𝐵𝐿𝑣
=
𝑅
𝑅
2
𝐵𝐿𝑣
Joule “heat” released in the wire: 𝑃 = 𝑅𝐼 2 =
𝑅
Who does the corresponding work? Of course, not the magnetic field! We do, by
pulling the loop with constant force:
𝐵2 𝐿2 𝑣
𝐵𝐿𝑣 2
𝐹𝑢𝑢 = 𝐼𝐼 𝐵 =
𝑃 = 𝐹𝐹 =
𝑅
𝑅
2. Lorentz force approach (we consider only electrons, assuming that the
force on the lattice is compensated by us):
𝑣
𝑣𝑑
- the horizontal segments
𝐹⃗𝐿
⃗
⃗
ℇ = � 𝐸𝑁𝑁 ∙ 𝑑𝑙 = �
∙ 𝑑𝑙 = 𝑣𝑣𝑣
𝜙
contribute nothing, since the force
𝑞
𝐹𝐿
is perpendicular to the wire)
𝑙𝑜𝑜𝑜
𝑙𝑜𝑜𝑜
For this kind of problems, the Lorentz force provides the so-called motional e.m.f., and
the Faraday’s law is just a short-cut which doesn’t contain any new physics.
11
Motional EMF (cont’d)
A metal disc rotates about a horizontal axis through a uniform
magnetic field B. A circuit is made by connecting one end of a
resistor to the axle and the other end to a sliding contact which
touches the outer edge of the disc. Find the current through
the resistor.
𝑎
𝐹⃗𝐿
𝐹𝐿 = 𝑞𝑞 𝐵
ℇ = � 𝐸𝑁𝑁
𝑙𝑜𝑜𝑜
- directed along the radius
𝑎
𝑎
𝐹⃗𝐿
𝑎2
∙ 𝑑𝑙⃗ = �
∙ 𝑑𝑙⃗ = � 𝑣𝑣𝑣𝑣 = � 𝜔𝑟𝐵𝐵𝐵 = 𝜔𝜔
𝑞
2
𝑙𝑜𝑜𝑜
ℇ 𝜔𝜔𝜔2
𝐼= =
2𝑅
𝑅
0
0
General expression for the motional e.m.f.:
ℇ = � 𝑣⃗ × 𝐵 ∙ 𝑑𝑙⃗
𝑙𝑜𝑜𝑜
Sometimes the motional e.m.f. can be successfully
described using both the Faraday’s law and the
approach based on the Lorentz force.
13
Loop Rotating in Magnetic Field
𝜙 = 𝜔𝜔
Φ𝑚
= � 𝐵 ∙ 𝑑𝐴⃗ = 𝐵𝐵 𝑐𝑐𝑐 𝜔𝑡
𝑑Φ𝑚
ℇ=−
= 𝜔𝐵𝐵 𝑠𝑠𝑠 𝜔𝑡
𝑑𝑑
14
Alternator / DC Generator
Alternator: generator of AC current.
DC generator: generator of DC current (DC motor in reverse).
15
Eddy Current (Foucault) Brake
Disc rotates through a B field → eddy
currents.
Eddy currents → new B field (Beddy)
Beddy opposes the rotation (Lenz!)
Linear eddy current brake
(useless at low speed, efficient at high speed)
The current direction in
the coils positioned along
the rail is alternating:
17
Conclusion
ℇ = � 𝐸𝑁𝑁 ∙ 𝑑𝑙⃗ = −
𝑙𝑜𝑜𝑜
𝑑Φ𝐵
𝑑𝑑
Faraday’s Law (the 3d Maxwell’s Eq.)
Motional e.m.f. – due to Lorentz force. Depending on the experiment set-up, can also
sometimes be treated on the basis of Faraday’s law.
Explanations are reference-frame dependent!
Next time: Lecture 19. Maxwell’s Equations
29.7
18