Download Brock University Physics 1P22/1P92 Winter 2014 Dr. D`Agostino

Brock University
Physics 1P22/1P92
Winter 2014
Dr. D’Agostino
Solutions for Tutorial 9: EM Induction and EM Waves (Chapter 25)
1. Each figure shows one or more metal wires sliding on fixed, conducting rails in a
magnetic field. Determine whether the induced current in each case is clockwise,
counter-clockwise, or zero in each case. [6 points]
Solution: (a) The size of the loop increases, so the magnetic flux through the loop
increases, with the magnetic field directed out of the page. Therefore the magnetic
field produced by the induced current points into the page, to oppose the change in
magnetic flux. Thus, the induced current is clockwise.
(b) The size of the loop increases, so the magnetic flux through the loop increases,
with the magnetic field directed into the page. Therefore the magnetic field produced
by the induced current points out of the page, to oppose the change in magnetic flux.
Thus, the induced current is counter-clockwise.
(c) The magnetic field has zero component parallel to the loop’s axis, so the magnetic
flux through the loop is zero. Thus, the change in magnetic flux is also zero, so the
induced current is zero.
(d) The size of the loop decreases, so the magnetic flux through the loop decreases,
with the magnetic field directed out of the page. Therefore the magnetic field produced
by the induced current points out of the page, to oppose the change in magnetic flux.
Thus, the induced current is counter-clockwise.
(e) The size of the loop does not change, nor does its orientation, nor does the magnetic
field. Thus, the magnetic flux through the loop does not change, and so the induced
current is zero.
(f) The wire moves in a region where the magnetic field is zero. Thus the magnetic
flux through the loop does not change as the wire moves, and so the induced current
is zero.
2. The loop in the figure is being pushed into the uniform 0.20 T magnetic field at a
speed of 50 m/s. The resistance of the loop is 0.10 Ω. Determine the magnitude and
direction of the current in the loop. [4 points]
Solution: First use Faraday’s law to determine the magnitude of the induced potential
difference in the loop:
∆Φ
∆t
∆ (AB cos θ)
=
∆t
∆A
= B cos θ
∆t
∆ (wL)
= B cos θ
∆t
∆w
= BL cos θ
∆t
= BLv cos θ
= (0.20 T) (0.05 m/s) (50 m/s) (cos 0◦ )
= 0.50 V
∆V =
∆V
∆V
∆V
∆V
∆V
∆V
∆V
Using Ohm’s law, we can determine the magnitude of the induced current:
∆V
R
0.50 V
I=
0.10 Ω
I = 5.0 A
I=
Finally, use Lenz’s law to determine the direction of the induced current. The magnetic
flux through the loop is increasing, so the induced current produces a magnetic field
that opposes the external magnetic field. Thus, the induced current is clockwise.
3. A helium-neon laser emits light in a 1.0-mm diameter beam of wavelength 633 nm
with power 1.0 mW. Determine the amplitudes of the electric and magnetic fields in
the laser beam. [4 points]
Solution: The intensity of the laser light is
P
A
P
I= 2
πr
1.0 × 10−3 W
I=
π (0.5 × 10−3 m)2
I = 1.273 × 103 W/m2
I=
The amplitude of the magnetic field satisfies
I=
1 c 2
· B
2 µ0 0
and therefore
2µ0 I
rc
2µ0 I
B0 =
c
B0 = 3.3 × 10−6 T
B02 =
The amplitude of the electric field is therefore
E0 = cB0
E0 = 9.8 × 102 V/m