Download Lecture 19: Motional emf

Phy2005
Applied Physics II
Spring 2017
Announcements:
Last Time
(& time before last)
Faraday’s law: any time there is a time-changing magnetic flux through
an area, there is an electromotive force (voltage) tending to drive
current around the boundary of the area
Vind = - A (dB/dt) rate of ch of B with time
= - d/dt (rate of ch of  with time)
 = BA= B A : magnetic flux
Sign in Faraday’s law – “Lenz’s law”
Current flows in a direction so as to oppose
the change in flux it experiences
Last time: practice w/ Faraday
Magnet always feels resistance to its motion.
Demos
Earth field generator
Faraday’s law with diode
Q1. ( 23.7) A flat loop of wire has an
area of 0.004 m2. It is in a region where
B=0.02 T and directed along the x axis.
 is the angle between the normal of the loop
and the x axis. The loop is rotated from
=0o to =60o in 0.5s. What is the change in flux?
d = BfA-BiA = BAcos f-BAcos i)
= (0.02)(0.004)(1/2 – 1)
= -4 x 10-5 T-m2
A. 0
B. -1.1×10-4 V
C. 8 x 10-5 T
D. -4 x 10-5 T-m2
E. 4 x 10-1 T-m2
B

B

B
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Q2. ( 23.7) A flat loop of wire has an
area of 0.004 m2. It is in a region where
B=0.02 T and directed along the x axis.
 is the angle between the normal of the loop
and the x axis. The loop is rotated from
=0o to =60o in 0.5s. What is the induced EMF?
A. 0
B. 1.1×10-4 V
C. 8 x 10-5 V
D. 4 x 10-5 V
E. 1.1 x 10-1 V

B
Ex 18-3 A circular loop with a 10 cm-radius is placed in the
presence of a uniform magnetic as shown in the figure. The
field changes from 1.5 T to 0.5 T in 0.5 s. The loop has
resistance 10W. What is the induced EMF? What current
flows (specify the direction and magnitude)?
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Vind = -d/dt
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d = f - I = -0.031 Tm2
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Vind = 0.062 V
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I = Vind/R = 0.0062 Amp
Field B out of screen
Motional emf
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i = LdB
V dt
f = L(d+vdt)B
Vind = - d/dt
= - LvdtB/dt
= - LvB
One way to think about it: electrons
In wire experience Lorentz force downwards
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B
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Current flows counterclockwise; electrons flow clockwise
Vind = - B dA/dt = B (vdt)L/dt = BvL
L
23-30 HW challenge: sliding
wire (no friction) on inclined rails.
Rails are resistanceless, but
wire has resistance R. What is
the terminal velocity?
Q1:Observation: wire slides down, accelerating until it
reaches a terminal velocity. Why????
A1: Because induced emf causes counterclockwise current,
sliding wire experiences increasing force component up the
rails, counteracting gravity
Q2: Total mechanical energy not conserved, since KE is const
and PE decreasing --- where is energy going?
A2: into heat dissipated in the resistor
Newton’s 2nd law: balance forces
Iind
Fwire
mg
Fwire
a
Vind
lvB cos a
l 2 B 2v cos a
 I ind lB  (
)(
)lB 
R
R
R
mgR sin a
Fwire cos a  mg sin a  v  2 2
l B cos 2 a