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Problem 1
Part A
A long straight wire carries a current ๐ผ.
(a) Consider a short segment โ๐ along the wire. The current flowing through this segment
generates a magnetic field around the wire. Explain how the Biot-Savart law can be used to
calculate the magnetic field at some arbitrary point P away from the wire. It is not necessary
to perform the calculation โ an explanation of the key parameters is all that is required. A
diagram would aid in your explanation
Solution:
The Biot-Savart law measures the magnetic field of a moving charge and is dependent on the
velocity, charge and displacement.
๐๐๐๐๐ก ๐โ๐๐๐๐
0
2
A charge has velocity for a straight wire:
Rearranging the equation, we get the Biot-Savart law for the magnetic field of a very short
segment of wire with carrying current ๐ผ:
๐๐ข๐๐๐๐๐ก ๐ ๐๐๐๐๐๐ก
0
๐ผโ๐ ร ๐ฬ
2
where ๐
0
= ๐๐๐๐๐๐ก๐๐ ๐๐๐๐๐๐๐๐๐๐๐ก๐ฆ, ๐ = ๐๐๐ ๐ก๐๐๐๐, ๐ผ = ๐๐ข๐๐๐๐๐ก and โ๐ =
(b) Although the Biot-Savart law can be used to calculate the magnetic field of a current-
carrying wire, a simpler way is to use Ampereโs law. Derive the equation for the magnetic
field at a distance d from the wire using Ampereโs law
Solution
Deriving Ampereโs Law:
We can see the integration is operated on a closed curve and evaluate that ๐ต
is tangent to a
line of length ๐ and a constant and equal magnitude everywhere along the line:
From this, we can observe that path length ๐ is the circumference of the wire which in this
case is the circumference of a circle, 2 ๐๐:
Therefore, the derivation for the equation is:
0
(c) Suppose that the current in the wire is increasing at a constant rate, and that the wire is
placed near (but not in electrical contact with) a conducting loop of wire with a resistance R.
The figure immediately below shows three such scenarios. For each case, discuss whether a
current would be induced in the loop by the changing current in the straight wire. If a
current is induced, then give the direction in which it would flow. If not, explain why not.
i) When the wire is pointing into the loop, the magnetic field is increasing and due to the
right-hand rule, the current will point in the cw direction. However, due to Lenzโs law,
the current in the loop must oppose this change. Therefore, the induced current is in the
counter-clockwise direction.
ii) There is no applied magnetic field given that the field does not pass through the loop. As
such, there is no induced current.
iii) When the wire is placed near the conducting loop, the magnetic field is increasing and
due to the right-hand rule, the magnetic field will point out of the page. However, due to
Lenzโs law, the current in the loop must oppose this change. Therefore, the induced
current is in the counter-clockwise direction.
The force acting on both the top and bottom horizontal sections are equal to:
2
๐๐ ร ๐ต
0
1
2
๐๐
2 ๐
0
1
2
[
ln(๐ฅ)
]
2 ๐
๐๐
0
1
2
ln (
Therefore, the net force is:
ln
0
1
2
0
1
2
ln (
๐๐๐ก
0
1
2
(ln( 2 ) + ln (
Using log rules:
๐๐๐ก
0
1
2
ln
Where ๐ is the number of times ๐ length of the wire is repeated for the horizontal segment.
Problem 2
Part A
Maxwellโs equations can be used to show that electromagnetic waves can propagate through space
(a) Describe the key aspects of an electromagnetic wave. Your description should mention
the electric and magnetic fields, direction of propagation, and speed. A diagram would
be useful in explaining these concepts
Solution
Electromagnetic waves are self-sustaining oscillations of electric and magnetic fields
which propagate through space without the charge or currents. The key aspects which
define an electromagnetic wave are that:
- The electric and magnetic fields oscillate and are perpendicular to the direction of
propagation ๐ฃ
- Both electric and magnetic fields are perpendicular to each other
- Regardless of frequency or wavelength, the velocity of all electromagnetic waves is
3 ร 10
8
(speed of light)
- The field strengths at every point in the wave are bound by the equation:
(b) At some point in space, a sinusoidal electromagnetic wave has an intensity of
โ 2
. Calculate the amplitudes of the electric field and the magnetic field at this
point. Ensure that you include units
Solution
To calculate the energy flow of an electromagnetic wave is to use the Poynting vector ๐,
defined by the formula:
0
2
0
โ 2
Following the first polarizer comes a second with some angle ๐ where ๐ผ
๐๐ข๐ก
โ 2
, therefore:
0
cos
2
Using trigonometric identities:
0
cos( 2 ๐)
cos( 2 ๐)
- 36 = cos( 2 ๐)
cos
โ 1
Therefore, the angle between the polarizers is 34 .45ยฐ.
Part B (advanced)
A double slit interference pattern is observed on a screen 1.0 m behind the slit. The separation of
the slits is 0.25 mm, and the width of each slit is much smaller than the separation. The slit is
simultaneously illuminated with two laser light sources โ a green laser with wavelength 532 nm and
a blue laser with wavelength 355 nm. Describe the pattern observed on the screen. Are there any
points where light from only one of the two lasers is observed? If so, calculate the position(s). If not,
explain why not.
Solution
For youngโs double slit experiment, the position of of bright fringes can be formulated by the
equation;
Where, ๐ = ๐๐๐ ๐๐ก๐๐๐ ๐๐ ๐๐๐๐โ๐ก ๐๐๐๐๐๐๐ ( 0 , 1 , 2 , 3 โฆ ), ๐ = ๐ค๐๐ฃ๐๐๐๐๐๐กโ ๐๐ ๐๐๐โ๐ก, ๐ฟ =
Likewise, for dark fringes, the equation can be formulated by adding ยฝ to m which yields:
The expected pattern on the screen when performed simultaneously are two patterns of blue and
green alternating bright and dark fringes which overlap each other causing interference. For blue,
given the smaller wavelength of 355nm, will result in the fringe spacing to be narrower than green
light which has a larger wavelength of 532nm.
As for the position when one colour from one laser source is visible, this would require the bright
fringe of one source and the dark fringe of the other source to equal each other on some fringe.
1
2
This is possible for youngโs double slit experiment given that the superposition of green and blue will
yield a blue and green pattern with an overlap of cyan (green and blue).
Problem 3
Part A
A copper wire with resistance 0.010 ฮฉ is shaped into a complete circle of radius R = 10 cm and
placed in a long solenoid so that the axis of the solenoid and the axis of the wire loop coincide. The
current in the solenoid is turned on and then slowly decreased. The magnetic field strength is
initially B = 0.750 T and subsequently decreases in time at the constant rate - 0.035 T/s.
(a) Calculate the induced emf in the ring while the magnetic field is decreasing.
Solution
If we relate back to Faradayโs Law, which states the emf induced around a closed loop if the
magnetic flux through the loop changes is equal to:
Given that the magnetic field decreases at a constant rate of โ 0. 035 ๐/๐ , and the radius is
10cm, we can find the emf:
2
ร |
๐B
2
ร 0. 035
Part B (Advanced)
Suppose that, instead of forming a complete ring, the two ends of the wire are connected to the
electrodes of a parallel-plate capacitor. The capacitor plates are circular with radius 1.0 cm and
separation 1.0 mm. Again, the magnetic field strength is initially B = 0.750 T and subsequently
decreases in time at the constant rate - 0.035 T/s.
(a) Sketch a charge diagram illustrating the final charge distribution on the capacitor plates. When
this final distribution is obtained, what is the total electric field in between the plates of the
capacitor?
Solution
(b) Explain why there is an induced magnetic field between the plates while the capacitor is
charging
Solution
A magnetic field which is created by a changing electric field is called an induced magnetic field.
This is caused by an increasing solenoid current due to charging as there is movement of charge
and current. This is formerly known as the displacement current which โallowsโ current to be
continuous through the capacitor and creates the same magnetic field with a changing electric
flux.
(c) Sketch the magnitude of the induced magnetic field as a function of the distance r from the axis
of the plates, giving values for r = 0.5 cm and r = 1.0 cm
Solution
+Q
- Q
