32: Magnetic Field from Two Wires

Figure 1

Figure 2

INTRO:
From the Biot-Savart law, it can be calculated that the magnitude of the magnetic field due to a long straight wire is given by
B_wire = μ₀I / 2πd ,

where μ0 (=4π×10⁻⁷ T⋅m/A) is the permeability constant, I is the current in the wire, and d is the distance from the wire to the location at which the magnitude of the magnetic field is being calculated.

The same result can be obtained from Ampere's law as well.

The direction of vector B⃗ can be found using the right-hand rule.

In this problem, you will be asked to calculate the magnetic field due to a set of two wires with antiparallel currents as shown in the diagram (figure 1). Each of the wires carries a current of magnitude I. The current in wire 1 is directed out of the page and that in wire 2 is directed into the page. The distance between the wires is 2d. The x-axis is perpendicular to the line connecting the wires and is equidistant from the wires.

As you answer the questions posed here, try to look for a pattern in your answers.

PART A:
Which of the vectors best represents the direction of the magnetic field created at point K (figure 1) by wire 1 alone?
Enter the number of the vector with the appropriate direction (figure 2).

SOLUTION:
For this, begin by pointing your thumb towards yourself since the current in wire 1 is directed out of the page. This means that the magnetic field will follow the direction of your fingers, counterclockwise. At point K, the question asks, what is the direction of the magnetic field created by only wire 1? Wouldn't that just be the tangential direction of the circular magnetic field at that point? Yes, it would be pointing to the right, or to choose one of the given vectors, vector 3.

PART B:
Which of the vectors best represents the direction of the magnetic field created at point K (figure 1) by wire 2 alone?
Enter the number of the vector with the appropriate direction (figure 2).

SOLUTION:
Do the exact same thing you did for the last problem except for this time point your thumb into the page, resulting in your fingers curling clockwise. At point K, the tangent of this circle would be pointing to the right. This means that vector 3 is our answer again. 

PART C:
Which of these vectors best represents the direction of the net magnetic field created at point K (figure 1) by both wires?
Enter the number of the vector with the appropriate direction (figure 2).

SOLUTION:
Think about the two previous answers. Since both of them were pointing to the right, vector 3, the only logical answer is that the net magnetic field is also pointing in the direction of vector 3.

PART D:
Find the magnitude of the magnetic field B_1K created at point K by wire 1.
Express your answer in terms of I, d, and appropriate constants.

SOLUTION: 
Since all of the values are already in the formula, B_wire = μ₀I / 2πd, we don't have to do anything else. The answer is μ₀I / 2πd
keep in mind that since K is equidistant from both wires, B_1K = B_2K

PART E: 
Find the magnitude of the net magnetic field B_K created at point K by both wires.
Express your answer in terms of I, d, and appropriate constants.

SOLUTION:
B_K = B_1K + B_2K
[μ₀I / 2πd] + [μ₀I / 2πd] = μ₀I / πd

PART F:
Point L is located a distance d(√2) from the midpoint between the two wires. Find the magnitude of the magnetic field B_1L created at point L by wire 1.
Express your answer in terms of I, d, and appropriate constants.

SOLUTION: 
The first thing we must do in this problem is determine the distance between point L and either one of the wires (equidistant again). We do this by setting up a right triangle and solving for the hypotenuse. Using Pythagorean's theorem, d²+[d(√2)]² = r²
r=√(d²+2d²) = √(3d²) = d(√3)
the magnetic field at this point L would be μ₀I / [2πd(√3)]

PART G:
Point L is located a distance d(√2) from the midpoint between the two wires. Find the magnitude of the magnetic field B_L created at point L by both wires.
Express your answer in terms of I, d, and appropriate constants.

SOLUTION: 
We can use the value from the previous part. Do not forget that this value is a vector, so we must take the x and y components into consideration if we are to determine the correct net magnetic field. We can set up a proportional right triangle using the ratio of the distances. The hypotenuse would be √3, the adjacent component would be 1 and the opposite component would be √2
y_1L = B_1L⋅sin(θ) = μ₀I⋅ 1/[2πd(√3)]⋅[√2/√3]= μ₀I⋅(√2)/ [6πd]
x_1L = B_1L⋅cos(θ) = μ₀I⋅ 1/[2πd(√3)]⋅1/(√3) = μ₀I/[6πd]
y_2L = B_2L⋅sin(-θ) = μ₀I⋅ 1/[2πd(√3)]⋅[√2/√3]= -μ₀I⋅(√2)/ [6πd]
x_2L = B_2L⋅cos(-θ) = μ₀I⋅ 1/[2πd(√3)]⋅1/(√3) = μ₀I/ [6πd]
Now that we have the x and y components, we are able to determine the magnetic field at point L
B_x = x_1L + x_2L = μ₀I/[6πd] + μ₀I/[6πd] = μ₀I/[3πd]
B_y = y_1L + y_2L = μ₀I⋅(√2)/ [6πd] + -μ₀I⋅(√2)/ [6πd] = 0

This leaves us with the magnitude of the vector <μ₀I/[3πd], 0>
which is μ₀I/[3πd]

Chapter 26: Overview

CHAPTER 25: THE ELECTRIC FIELD

Coulomb’s law:

F = K⋅|q₁|⋅|q₂| / r²

Electric field of a point charge
E = K ⋅ q/r² ⋅ (r^)

Useful:

Electric constant
ε₀ = 8.854×10^-12 F/m

SI Unit conversions
1 N = 1 kg⋅m⋅s^-2
1 C = A⋅s
1 F = A²⋅s⁴⋅kg^-1⋅m^-2

25: Problem 25.66

INTRO:
An electric field E=100,000 ı^ N/C causes the 5.0 g point charge in the figure to hang at a 20° angle.

Figure 1

Part A: 
What is the charge on the ball?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
q = 180 nC

25: Problem 25.59

INTRO:
Two 3.40g point charges on 1.00-m-long threads repel each other after being equally charged, as shown in the figure.Assume that θ = 28

Part A: 
What is the charge q?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
q = -1.32 μC

25: Problem 25.57

INTRO:
You sometimes create a spark when you touch a doorknob after shuffling your feet on a carpet. Why? The air always has a few free electrons that have been kicked out of atoms by cosmic rays. If an electric field is present, a free electron is accelerated until it collides with an air molecule. It will transfer its kinetic energy to the molecule, then accelerate, then collide, then accelerate, collide, and so on. If the electron’s kinetic energy just before a collision is 1.80×10⁻¹⁸J or more, it has sufficient energy to kick an electron out of the molecule it hits. Where there was one free electron, now there are two! Each of these can then accelerate, hit a molecule, and kick out another electron. Then there will be four free electrons. In other words, as shows below, a sufficiently strong electric field causes a “chain reaction” of electron production. This is called a breakdown of the air. The current of moving electrons is what gives you the shock, and a spark is generated when the electrons recombine with the positive ions and give off excess energy as a burst of light.

Part A: 
The average distance an electron travels between collisions is 2.00 μm . What acceleration must an electron have to gain 1.80×10⁻¹⁸J of kinetic energy in this distance?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
a = 9.88×10¹⁷ m/s²

Part B:
What force must act on an electron to give it the acceleration found in part A?
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
F = 9.00×10^-13 N

Part C:
What strength electric field will exert this much force on an electron? This is the breakdown field strength. 
Note: The measured breakdown field strength is a little less than your calculated value because our model of the process is a bit too simple. Even so, your calculated value is close.
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
E = 5.63×10⁶ N/C

Part D:
Suppose a free electron in air is 0.700 cm away from a point charge. What minimum charge must this point charge have to cause a breakdown of the air and create a spark? Assume the electron is not hitting the charge particle but a neutral air atom that is still 2.00 μm away.
Express your answer to three significant figures and include the appropriate units.

SOLUTION:
q = 30.6 nC

25: Problem 25.51

INTRO:
In a simple model of the hydrogen atom, the electron moves in a circular orbit of radius 0.053 nm around a stationary proton.

Part A: 
How many revolutions per second does the electron make?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
6.6×10¹⁵ rev/s

UPDATE, 09/12/2016

The first thing I always do is convert my givens to SI units 

r = 0.053 nm = 5.3×10^-2 nm = 5.3×10^-11 m

Now. We know an ELECTRON is moving around a PROTON, so we can look up the charges for these two items

q₁ = q_PROTON = 1.602×10^-19 C
q₂ = q_ELECTRON = -1.602×10^-19 C

Next we need to do a basic dynamics problem. 

There is an electromagnetic force pulling the electron around the neutron. F = ma
the acceleration is tangential to the orbit, so a = a_n = ω²⋅r
ω = v/r ∴ ω² = v² / r² ∴ a_n = v² / r² ⋅ r = v² / r
⇒ F = m⋅a_n = m⋅v² / r

Since we're trying to find the velocity and know everything else except the force (and this entire section has to do with Force & Coulomb's Law...)

F = k_e⋅q₁⋅q₂/r²
∴ k_e⋅q₁⋅q₂/r² = m⋅v² / r

⇒ k_e⋅q₁⋅q₂ / (r⋅m) = v²
∴ v = (k_e⋅q₁⋅q₂ / (r⋅m) )^½

​

The mass of an electron is m_e = 9.109×10^-31 kg
v = (k_e⋅q₁⋅q₂ / (r⋅m) )^½ 

= SQRT{ (8.99×10⁹ Nm²/C²) ⋅ (1.602×10^-19 C) ⋅ (-1.602×10^-19 C) ⋅ 1/(9.109×10^-31 kg) ⋅ 1/(5.3×10^-11 m) ⋅ (kg⋅m/s²)/N }

= SQRT{ -4.779 × 10¹² m² / s² }

--> v = 2.186 × 10⁶ m/s

convert to rev/s -> 1 rev / (2 π r) = 1 rev / 3.33×10^-10 m

v = 6.565 rev/s

*update (07/26/2017 @ Kirtan Jani) * v = 6.565×10¹⁵ rev/s =  6.6×10¹⁵ rev/s

This was wrong because I missed an exponent while explaining. However, if you look at the purple highlighted answer above.... you can see that that was only wrong because you were being too GD judgemental.  Remember this is a free service bruh.. 

25: Problem 25.40

Part A: 
What is the force F⃗ on the 1 nC charge at the bottom?
Write your answer as two vector components, separated by a comma. Express each component numerically, in newtons, to two significant figures.

SOLUTION:
F_x, F_y = 0, 1.1×10^-5 N

25: Problem 25.33

Part A: 
What is the magnitude of the force F⃗ on the 1.0 nC charge in the figure?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
F = 3.1×10^-4 N

Here's why ->

Part B:
What is the direction of the force F⃗ on the 1.0 nC charge in the figure?

SOLUTION:
Upward

25: Problem 25.9

INTRO:
Two neutral metal spheres on wood stands are touching. A negatively charged rod is held directly above the top of the left sphere, not quite touching it. While the rod is there, the right sphere is moved so that the spheres no longer touch. Then the rod is withdrawn.

Part A: 
Afterward, what is the charge state of each sphere?

a) Both the spheres are neutral.
b) The left sphere is negatively charged, the right sphere is charged positively.
c) The right sphere is negatively charged, the left sphere is charged positively.
d) Both the spheres are charged positively.
e) Both the spheres are charged negatively.

SOLUTION:
option c)

25: Problem 25.30

INTRO:
The nucleus of a ¹²⁵Xe atom (an isotope of the element xenon with mass 125 u) is 6.0 fm in diameter. It has 54 protons and charge q = +54e.

Part A: 
What is the electric force on a proton 1.4 fm from the surface of the nucleus? 
Hint: Treat the spherical nucleus as a point charge.
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
We know the formula for the force between two charged particles is F = K |q₁| |q₂| r^-2
We simply have to determine which values to plug into this equation and we will get the answer that we're looking for. 
Begin by taking the radius of the nucleus, which is half the diameter, or 3 fm
the added distance of the proton is 1.4 fm, so the total distance of the proton from the center of the nucleus is 3 + 1.4 = 4.4 fm = 4.4⋅10^-15 m

so r = 4.4⋅10^-15 m​

Next we need to determine the two charges. 
We're given that the nucleus has a charge of q = +54e. 
Knowing that e=1.6⋅10^-19 C , q_nucleus = 8.64⋅10^-18 C

q₁ = 8.64⋅10^-18 C​

lastly, we know that the charge of ONE proton is just e, so

q₂ = 1.6⋅10^-19 C​

Now we have all of our values to plug in and can solve for the force
(remember K is the electrostatic constant and equal to 9.0⋅10⁹ Nm²/C²)

∴F = (9.0⋅10⁹ Nm²/C²)(8.64⋅10^-18 C)(1.6⋅10^-19 C)(4.4⋅10^-15 m)^-2​

→F = 642.645 N​

F = 643 N

Part B:
What is the proton's acceleration?
Express your answer to two significant figures and include the appropriate units.

SOLUTION:
We know that F=ma, so if we divide the force by the given mass...  a = 3.8×10²⁹ m/s²