34: Enhanced EOC: Problem 34.14

INTRO:
The magnetic field of an electromagnetic wave in a vacuum is B_z=(2.0μT)sin((9.00×10⁶)x−ωt), where x is in m and t is in s.

Part A:
What is the wave's wavelength?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
a value for k is given in the formula
k=9.00×10⁶
since λ=2π⋅1/k, 
λ = 2π⋅1/9.00×10⁶ = 6.98×10^-7 m= 698 nm

Part B:
What is the wave's frequency?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
using the formula c=fλ, we can easily solve for frequency 
f = c/λ = 3.0×10⁸m / 6.98×10^-7 m⋅s
f = 4.3×10¹⁴ Hz

Part C:
What is the wave's electric field amplitude?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
E₀=cB₀
→c(2.0μT) 
E₀ = 599.6 N/C ≈ 600 V/m

34: Electric and Magnetic Field Vectors Conceptual Question

INTRO:

Figure 1

Figure 2

Figure 4

Figure 3

all of the questions will refer to the following answer choices:

a) +x
b) –x
c) +y
d) –y
e) +z
f) –z
g) at a +45∘ angle in the xy plane

Part C:
The electric and magnetic field vectors at a specific point in space and time are illustrated. (Figure 1) Based on this information, in what direction does the electromagnetic wave propagate?

SOLUTION:
the Poynting Vector S points in the direction of E×B
which in this instance would point in the positive z direction, or option e)
~ thumb is E, pointer is B → middle finger points +z

Part B:
The electric and magnetic field vectors at a specific point in space and time are illustrated. (Figure 2) (E and B are in the xy plane. Both vectors make45∘ angles with the +y axis.) Based on this information, in what direction does the electromagnetic wave propagate?

SOLUTION:
Since E & B are both in the xy plane, their cross product will be along the z-axis
In this case, E×B points in a negative z direction, or option f)
~thumb is E, pointer is B, middle points -z

Part C:
The magnetic field vector and the direction of propagation of an electromagnetic wave are illustrated. (Figure 3) Based on this information, in what direction does the electric field vector point?

SOLUTION:
Basically just work backwards
~middle is v, pointer is B, thumb points +y → option c)

Part D: 
The electric field vector and the direction of propagation of an electromagnetic wave are illustrated. (Figure 4) (E is in xz plane and makes a 45° angle with the +x axis.) Based on this information, in what direction does the magnetic field vector point?

SOLUTION:
work backwards again but be mindful of angles
~thumb is E, middle is v, pointer points at 45° between +x & -z, or option g)

34: Problem 34.22

Part A:
At what distance from a 15W point source of electromagnetic waves is the magnetic field amplitude 0.60μT ?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
givens:
P_source = 15 W
B₀ = 0.60μT
and we're trying to determine the distance from the point source, r

How can we relate the magnetic field amplitude and the distance from the point source? 
neither of our basic field formulas helps us, but we do know a series of equations relating the Electric field amplitude and electromagnetic wave intensity, and the amplitudes of the Electric field and the Magnetic field are related by the speed of light, we can arrange the following formulas to solve for r

Intensity relations~ 
I= P_source⋅1/(4πr²) = c⋅ε₀⋅E₀²⋅1/2
Magnetic & Electric field relations~
B₀⋅c = E₀
so E₀² = (B₀⋅c)² 
∴ I= c⋅ε₀⋅(B₀⋅c)²⋅1/2=c³⋅ε₀⋅(B₀)²⋅1/2 
since I also equals P_source⋅1/(4πr²), it can be simplified that
P_source⋅1/(4πr²) = c³⋅ε₀⋅(B₀)²⋅1/2
now, ε₀ = 1/(μ₀c²) = 1/[(4π⋅10^-7 H/m)c²]
so P_source⋅1/(4πr²) = c³⋅1/[(4π⋅10^-7 H/m)c²]⋅(B₀)²⋅1/2 simplifies to
P_source⋅[2⋅10^-7 H/m] / (r²) = c⋅(B₀)² which simplifies to
r²=P_source⋅(2⋅10^-7 H) ⋅1/[c⋅B₀² ⋅m]
leaving r = √(P_source⋅(2⋅10^-7 H) ⋅1/[c⋅B₀² ⋅m])
now, plugging in for all of the values for which we've arranged so nicely, we get
r = √{(15 W)⋅(2⋅10^-7 H)⋅1/[(3⋅10⁸ m/s)(3.6⋅10^-13)⋅T)² ⋅m] }
→ the first thing I'm going to do here is solve for the units & check that I'm on the right path
√{W⋅H⋅1/(m/s)⋅1/(T²)⋅1/m}
W = kg⋅m⋅m⋅1/s⋅1/s⋅1/s & H = kg⋅m⋅m⋅1/s⋅1/s⋅1/A⋅1/A & T = kg⋅1/A⋅1/s⋅1/s →1/T = A⋅s⋅s⋅1/kg
⇒√{kg⋅m⋅m⋅1/s⋅1/s⋅1/s⋅kg⋅m⋅m⋅1/s⋅1/s⋅1/A⋅1/A⋅A⋅s⋅s⋅1/kg⋅A⋅s⋅s⋅1/kg⋅1/m⋅s⋅1/m}
⇒√{kg²⋅m⁴⋅A²⋅s⁵⋅1/kg²⋅1/m²⋅1/s⁵⋅1/A²}⇒√{m²}
⇒units=m
which is right so that means our equation is set up properly, now solve
√{(15)⋅(2⋅10^-7)⋅1/[1.08⋅10^-4] = √{1/36} = 1/6 = 0.17 m

34: Traveling Electromagnetic Wave

INTRO:
Light, radiant heat (infrared radiation), X rays, and radio waves are all examples of traveling electromagnetic waves. Electromagnetic waves comprise combinations of electric and magnetic fields that are mutually compatible in the sense that the changes in one generate the other.
The simplest form of a traveling electromagnetic wave is a plane wave. For a wave traveling in the x direction whose electric field is in the y direction, the electric and magnetic fields are given by

E =E₀sin(kx−ωt)j^
B =B₀sin(kx−ωt)k^

This wave is linearly polarized in the y direction.

Part A:
In these formulas, it is useful to understand which variables are parameters that specify the nature of the wave. The variables E₀ and B₀ are the __________ of the electric and magnetic fields.
Choose the best answer to fill in the blank.

A) maxima
B) amplitudes
C) wavelengths
D) velocities

Part B:
The variable ω is called the __________ of the wave.
Choose the best answer to fill in the blank.

A) velocity
B) angular frequency
C) wavelength

Part C:
The variable k is called the __________ of the wave.
Choose the best answer to fill in the blank.

A) wavenumber 
B) wavelength
C) velocity
D) frequency

Part D:
What is the mathematical expression for the electric field at the point x=0, y=0, z at time t?

A) E = E₀sin(−ωt)j^
B) E = E₀sin(−ωt)k^
C) E = 0
D) E = E₀sin(kz−ωt)i^
E) E = E₀sin(kz−ωt)j^

Part E:
For a given wave, what are the physical variables to which the wave responds?

A) x only
B) t only
C) k only
D) ω only
E) x and t
F) x and k
G) ω and t
H) k and ω

Part F:
What is the wavelength λ of the wave described in the problem introduction?
Express the wavelength in terms of the other given variables and constants like π.

SOLUTION:
λ = 2π⋅1/k

Part G:
What is the period T of the wave described in the problem introduction?
Express the period of this wave in terms of ω and any constants

SOLUTION:
frequency = 1/period ↔ period = 1/frequency
f = 1/T ↔ T = 1/f
f = ω/[2π] → T = 2π⋅1/ω

Part H:
What is the velocity v of the wave described in the problem introduction?
Express the velocity in terms of quantities given in the introduction (such as ω and k) and any useful constants.

SOLUTION:
v = fλ & λ = 2π⋅1/k & f = ω/[2π] 
so... v = 2π⋅1/k⋅ω/[2π] = ω/k

32: Video Tutor: Magnet and Electron Beam

INTRO:
The figure shows the path of a charged particle moving in a magnetic field directed into the screen.

What is the particle’s charge?

A) Negative

B) Positive

C) Neutral

SOLUTION: 

to get the correct direction for the centripetal force, the charge must be negative

32: Interaction of a Current Loop with a Magnetic Field

INTRO: 
The effects due to the interaction of a current-carrying loop with a magnetic field have many applications, some as common as the electric motor. This problem illustrates the basic principles of this interaction.

Figure 1

Figure 2

Figure 3

Figure 4

Consider a current I that flows in a plane rectangular current loop with height a = 4.00 cm and horizontal sides b = 2.00 cm . (Figure 1) The loop is placed into a uniform magnetic field B⃗  in such a way that the sides of length a are perpendicular to B⃗ , and there is an angle θ between the sides of length b and B⃗ , as shown in the figures. (Figure 2)

PART A:

Will the interaction of the current through the loop with the magnetic field cause the loop to rotate?

  A) Yes, the net torque acting on the loop is negative and tends to rotate the loop in the direction of decreasing angle θ (clockwise).

  B) Yes, the net torque acting on the loop is positive and tends to rotate the loop in the direction of increasing angle θ (counterclockwise). 

  C) No, the net torque acting on the loop is zero and the loop is in equilibrium.

  D) No, the net force acting on the loop is zero and the loop is in equilibrium.

SOLUTION:
By using the right hand rule where the pointer finger indicates the magnetic field and the thumb represents the moving particle (aka direction of current), if you refer to figure 2, you can see that the force on both sides of the ring pull the ring out, increasing the angle. This observation is consistent with option B

PART B:

For parts B and C, the loop is initially positioned at θ=30°.

Assume that the current flowing into the loop is 0.500A . If the magnitude of the magnetic field is 0.300T , what is τ, the net torque about the vertical axis of the current loop due to the interaction of the current with the magnetic field?

Express your answer in newton-meters.

SOLUTION:

First off we have to determine the magnitude of the force on each vertical bar, which they will be inversely equivalent.
So, using the formula F = ILB, 
F = (0.5 A)(0.04 m)(0.3 T) = 6×10^-3 N
NOW, since the ring is skewed, we have to disregard the component of the force that is along the 'lever'. 
Using our handy right triangle, we can determine that the perpendicular component of the force will be
F₁ = F cos(θ) = 5.2×10^-3 N
Now, recalling from the beginning of physics, torque = F×l 
τ₁ = (5.2×10^-3 N) ⋅ (0.01 m) [we get this l from half the length of the top bar]
τ₁ = 5.2×10^-5 N⋅m
but this isn't our answer quite yet, remember they asked for the net torque. What we just found was the torque for the 'upper' half of the ring. To get our net torque, we must add the torque from the 'top' to the torque from the 'bottom'
luckily, the two torques are equal so we just multiply our previous value by 2 and determine that the net torque is 
1.04×10^-4 N⋅m

PART C:

What happens to the loop when it reaches the position for which θ=90∘, that is, when its horizontal sides of length b are perpendicular to B⃗  (see the figure)? (Figure 3)

A) The direction of rotation changes because the net torque acting on the loop causes the loop to rotate in a clockwise direction.

 B) The net torque acting on the loop is zero, but the loop continues to rotate in a counterclockwise direction.

 C) The net torque acting on the loop is zero; therefore it stops rotating.

 D) The net force acting on the loop is zero, so the loop must be in equilibrium.

SOLUTION:

so once the ring rotates to the 90° point, there will be no more force on the edges of it, however, there won't be any forces on the other side either, resulting in it to continue rotating until something stops it, which is satisfied by B).

PART D:

Now suppose that you change the initial angular position of the loop relative to B⃗ , and assume that the loop is placed in such a way that initially the angle between the sides of length b and B⃗  is θ=120, as shown in the figure. (Figure 4) Will the interaction of the current through the loop with the magnetic field cause the loop to rotate?

A) Yes, the net torque acting on the loop is negative and tends to rotate the loop in the direction of decreasing angle θ (clockwise).

B) Yes, the net torque acting on the loop is positive and tends to rotate the loop in the direction of increasing angle θ (counterclockwise).

C) No, the net torque acting on the loop is zero and the loop is in equilibrium.

D) No, the net force acting on the loop is zero and the loop is in equilibrium.

SOLUTION:

Same idea as in the first problem, except in the opposite direction. This means that there will be torque on the ring just in the other way now. So, A).

32: Video Tutor: Current-Carrying Wire in Magnetic Field

PART A:

The figure shows a wire that is connected to a power supply and suspended between the poles of a magnet. When the switch is closed, the wire deflects in the direction shown.

Which of the dashed boxes A–D represents the position of the north magnetic pole?

SOLUTION: 
As he explains in the video, if you point your right thumb towards yourself to represent the direction of the current, and then your pointer finger down to represent the magnetic field pointing from north to south, the middle finger will represent the direction that the wire is displaced. 
Working backward, pointing your right thumb toward yourself and middle finger down in the direction of deflection, your pointer finger points to the left. Since this represents the magnetic field from north to south, the dashed box that represents the north magnetic pole is therefore C.

32: Force Between Two Infinite, Parallel Wires

INTRO: 
You are given two infinite, parallel wires each carrying current I. The wires are separated by a distance d, and the current in the two wires is flowing in the same direction. This problem concerns the force per unit length between the wires.

PART A:
Is the force between the wires attractive or repulsive?

SOLUTION: 
The second right hand rule assists us in solving this problem. Since we have two parallel wires with equal current going the same direction, that vector represents each of the wires velocity vectors. By pointing our right thumb in this direction and using our pointer finger to represent the magnetic field (pointing back toward ourselves), at any point, the Force of the wire will be represented by our middle finger. Thinking of this for both wires, the force vectors are pointing towards each other, resulting in an attractive relationship. 

PART B:
What is the force per unit length F/L between the two wires?
Express your answer in terms of I, d, and constants such as μ₀ and π.

SOLUTION:
First some relevant formulas:
F_B=qvBsin(θ) = qv^→×B^→ 
on a long straight wire,
F=ILBsin(θ) = I^→L×B^→ where L is the length of the wire
So the magnetic field at wire 2 from the current in wire 1 will be B=μ₀I₁ ⋅ 1/[2πd]
The force on a length ΔL of wire 2 will be F=ΔLI₂×B
The force per unit length in terms of the currents will be,
F/ΔL = μ₀I₁I₂ ⋅ 1/[2πd]
since I₁ = I₂,
F/L = μ₀I² / [2πd]

PART C:
In the SI system, the unit of current, the ampere, is defined by this relationship using an apparatus called an Ampère balance. What would be the force per unit length of two infinitely long wires, separated by a distance 1m, if 1A of current were flowing through each of them?
Express your answer numerically in newtons per meter.

SOLUTION:
If I = 1 A, & d = 1 m, plugging these values into the formula from the previous section,
F/L = μ₀(1 A)² / [2π(1m)]
= μ₀/ [2π] ⋅ A²/m
μ₀ = 4π×10^-7 ⋅ T⋅m/A
→ 4π×10^-7 / [2π] ⋅A²/m ⋅T⋅m/A  →  2×10^-7 ⋅ T⋅A
1 T = 1 N / [A⋅m] →2×10^-7 ⋅ N/m

32: Tactics Box 32.1 Right-Hand Rule for Fields

Figure 1

Figure 2

INTRO:
TACTICS BOX 32.1 Right-hand rule for fields

1. Point your right thumb in the direction of the current.(Figure 1)
2. Curl your fingers around the wire to indicate a circle.
3. Your fingers point in the direction of the magnetic field lines around the wire.

PART A: The following sketches show a wire carrying a current I in the direction indicated. Which sketch correctly shows the magnetic field lines around the wire?
a)b)c)

SOLUTION:
First off, we know that the magnetic field can't be going two different directions, such as it is in b), which rules out that answer. Next we must use the right hand rule to determine the direction it goes. Pointing our right thumb downward, in the direction thatI points, our fingers wrap counterclockwise around the bar. Since a) demonstrates this, we know that the correct answer is a). 

PART B: The magnetic field lines around a current-carrying wire are shown in the figure. In what direction is the current flowing in the wire? (Figure 2)

SOLUTION: For this one, pay special attention to which of the blue lines are in front of the diagram of the rod and which ones are behind, because this determines the direction of the magnetic field and also how to determine the direction of the current. So, that being said, place a right hand in the air and curl your fingers in the direction of the field, towards you as if you're looking at your fingernails. This results in the thumb pointing to the right. This means that this is the direction that the current is flowing in the wire. So the final answer for this part is 'from left to right'