Discussion Question 10B
P212, Week 10
Faraday's Law: Rotating Loop
ω = 8 rad/s
y
R=4Ω
Consider a rectangular loop of
wire with width w, height h, and
total resistance R. The region of
space occupied by the wire loop
contains a uniform magnetic field
B pointing out of the page.
positive I
=
B = 1.2 T
h = 3 cm
clockwise
x
A motor attached to the loop
w = 5 cm
keeps it rotating about the vertical
axis with constant angular
velocity ω. At time t = 0, the loop is parallel to the paper (as shown in the figure).
(a) Plot the magnetic flux ΦB through the loop as a function of time. Find the maximum value
(amplitude) of the flux, and write it on your plot.
Φ max = 1.8 mT im 2
z
B
y
x
π
2ω
y
x
ωt
η̂
π
ω
3π
2ω
The Φ is easiest to see in the x-z plane. I show the view of an observer that is looking along the
ˆ is ⊥ to the x-y plane in a direction given by
axis of rotation (the y axis). The normal η
the right hand rule for our current loop. At t = 0 this lies along the -z axis. The sense of the rotation
ˆ rotating in the -x direction as shown. By trig η
ˆ = − sin ωt ˆx − cos ωt ˆz
has η
ˆ ) =B(area) ( ˆziη
ˆ ) = -B × area × cosωt = −( 1.2T )( 0.03 × 0.05 m 2 )cosωt
Φ = B i(area η
→ Φ = −1.8 ×10−3 cos ωt T im 2
t t
E max = 14.4 mV
(b) Plot the induced EMF E through the loop as a function of
time. Find the maximum value (amplitude) of the EMF, and write
it on your plot.
dφ
d
= − ( −1.8 ×10−3 cos ωt ) = −1.8 × 10−3 ω sin ωt
dt
dt
E = − (1.8 × 10−3 ) ( 8 ) sin ωt = −14.4 × 10−3 sin ωt V
E =−
π
2ω
π
ω
t
3π
2ω
(c) Plot the induced current I through the loop as a function of time. Find the maximum value
(amplitude) of the current, and write it on your plot.
Ιmax= 3.6 mA
E −14.4 × 10−3
I= =
sin ωt = −3.6 sinωt mA
R
4Ω
π
ω
π
2ω
3π
2ω
(d) Plot the torque τ provided by the motor to keep the loop rotating at a constant angular
velocity. Find the maximum value (amplitude) of the torque, and write it on your plot.
µ points along the plane normal.
z
τmag = µ × Β = I (area )Β sin (180 − ωt )
B
= I (area )Β sin ωt yˆ
x
y
ωt
µ
τmax=6.48µN•m
τmag = ( −3.6 × 10 −3 sinωt ) ( 0.05 × 0.03
)(1.2 ) sin ωt yˆ
τmag = −6.48 sin 2 ωt yˆ µNim
The motor must supply
τmotor = +6.48 sin 2 ωt yˆ µNi m
to cancel the magnetic torque and keep
ω constant τmotor is in the direction of motion
and hence the motor is supplying work.
π
2ω
π
ω
3π
2ω
2π
ω
t