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Solutions

Question 1 of 30Physics  Electrostatics, DC Circuits, Magnetism
1. Question
1. Two electrons are placed 2 x 10^{8} meters apart from each other. What will be the relative repulsive force if the electrons are moved so that they are 1 x 10^{8} meters apart? Assume all else remains constant.Correct / You marked this questionStep 1: Let’s recall our equation for Coulomb’s Law (i.e. for the force between two charges)
Step 2: Analyze Scenario 1 vs. Scenario 2
Note: we must realize the the distance between the charges is ^{1}/_{2} as great in Scenario 1 vs. Scenario 2
Scenario 1
Scenario 2
You can see the force between the two charges is 4 times as great in Scenario 2.
Incorrect / You marked this questionStep 1: Let’s recall our equation for Coulomb’s Law (i.e. for the force between two charges)
Step 2: Analyze Scenario 1 vs. Scenario 2
Note: we must realize the the distance between the charges is ^{1}/_{2} as great in Scenario 1 vs. Scenario 2
Scenario 1
Scenario 2
You can see the force between the two charges is 4 times as great in Scenario 2.

Question 2 of 30Physics  Electrostatics, DC Circuits, Magnetism
2. Question
2. Which of the following is equivalent to electric field:Correct / You marked this questionThis is a conceptual question.
An electric field is defined as the collection of force vectors emanating from an electric source charge which would exert forces on a positive test charge.
 For a positive (+) source charge, the electric field vector points away from the positive source charge
 For a negative () source charge, the electric field vector points toward the negative source charge
If we examine our electric field equation, we can see that electric field is the force per unit charge, and is measured in
Newtons / Coulomb.Incorrect / You marked this questionThis is a conceptual question.
An electric field is defined as the collection of force vectors emanating from an electric source charge which would exert forces on a positive test charge.
 For a positive (+) source charge, the electric field vector points away from the positive source charge
 For a negative () source charge, the electric field vector points toward the negative source charge
If we examine our electric field equation, we can see that electric field is the force per unit charge, and is measured in
Newtons / Coulomb. 
Question 3 of 30Physics  Electrostatics, DC Circuits, Magnetism
3. Question
3. Which of the following are units of electric field?Correct / You marked this questionThis is a conceptual question.
An electric field is defined as the collection of force vectors emanating from an electric source charge which would exert forces on a positive test charge.
 For a positive (+) source charge, the electric field vector points away from the positive source charge
 For a negative () source charge, the electric field vector points toward the negative source charge
If we examine our electric field equation, we can see that electric field is the force per unit charge, and is measured in
Newtons / Coulomb.However, electric field can also be expressed in units of Volts / Meter!
Volts are the standard unit for electric potential, a form of electromotive force. If we measure electric field in terms of V/m, we are describing the electromotive force per distance between charges.
Incorrect / You marked this questionThis is a conceptual question.
An electric field is defined as the collection of force vectors emanating from an electric source charge which would exert forces on a positive test charge.
 For a positive (+) source charge, the electric field vector points away from the positive source charge
 For a negative () source charge, the electric field vector points toward the negative source charge
If we examine our electric field equation, we can see that electric field is the force per unit charge, and is measured in
Newtons / Coulomb.However, electric field can also be expressed in units of Volts / Meter!
Volts are the standard unit for electric potential, a form of electromotive force. If we measure electric field in terms of V/m, we are describing the electromotive force per distance between charges.

Question 4 of 30Physics  Electrostatics, DC Circuits, Magnetism
4. Question
4. Electric potential is typically measured in units of:Correct / You marked this questionThis is a conceptual question.
Volts are the standard unit for electric potential, a form of electromotive force.
 A common analogy to help understand electric potential is the waterflow pipe analogy: You can think of voltage, i.e. the difference in electric potential, as the change in water pressure in a pipe.
Incorrect / You marked this questionThis is a conceptual question.
Volts are the standard unit for electric potential, a form of electromotive force.
 A common analogy to help understand electric potential is the waterflow pipe analogy: You can think of voltage, i.e. the difference in electric potential, as the change in water pressure in a pipe.

Question 5 of 30Physics  Electrostatics, DC Circuits, Magnetism
5. Question
5. Volts are equivalent to:Correct / You marked this questionThis is a conceptual question.
Volts are the standard unit of electric potential between two points. “Voltage” is the electric potential energy per unit charge,
i.e joules / coulomb.This may be easier to understand in the context of a battery. Chemical reactions in the battery cause electrons to accumulate at the negative terminal. The electrons then flow from the negative terminal (highest PE) to the positive terminal (lowest PE). The PE of electrons at the negative terminal is converted into other forms of energy, much in the way PE at the top of a waterfall could be converted into KE as the water descends.
The amount of PE is dependent on the number of electrons. The greater the charge density the greater the potential.
Incorrect / You marked this questionThis is a conceptual question.
Volts are the standard unit of electric potential between two points. “Voltage” is the electric potential energy per unit charge,
i.e joules / coulomb.This may be easier to understand in the context of a battery. Chemical reactions in the battery cause electrons to accumulate at the negative terminal. The electrons then flow from the negative terminal (highest PE) to the positive terminal (lowest PE). The PE of electrons at the negative terminal is converted into other forms of energy, much in the way PE at the top of a waterfall could be converted into KE as the water descends.
The amount of PE is dependent on the number of electrons. The greater the charge density the greater the potential.

Question 6 of 30Physics  Electrostatics, DC Circuits, Magnetism
6. Question
6. Three negative charges are located as shown in the following diagram. A charge of “e” corresponds to the charge of an electron, “2e” is twice the charge of an electron, and so on. Assume the three charges are equidistant from each other, and that each numbered location is equidistant from the charges.
We place a positive test charge into the system. At which location would the force on the test charge be strongest?Correct / You marked this questionStep 1: Describe how we determine the force on a test charge at the respective locations
For a negative source charge, a positive test charge will be attracted toward the negative source charge.
Thus, you can picture the force vectors for our diagram as follows:
Per Coulomb’s Law, the force on the test charge will be equal to:
We don’t actually need to solve for the force at each point. We know that the test charge is constant at any location. Thus, the force on the test charge caused by each source charge will be proportional to the magnitude of the source charge. That is, if q_{1} is the test charge, q_{2} is the source charge.
Thus, to determine the net force at any point on the diagram, we simply add up the vectors at that location.
Step 2: Perform vector addition at each location
 Point 1: the test charge is attracted to the right with a magnitude proportional to 3e. The other test charge distances rule those forces out.
 Point 2: the test charge is attracted to the left with a magnitude proportional to 3e, and to the right with a magnitude proportional to e. Thus, the net force has a magnitude less than (proportional to) 3e
 Point 3: the test charge is attracted to the left with a magnitude proportional to e, and to the right with a magnitude proportional to 2e. Thus, the net force has a magnitude less than (proportional to) 2e
 Point 4: the test charge is to the left with a magnitude proportional to 2e. Just like for point 1, the other test charge distances rule those forces out.
As you can see, the force on the test charge is greatest at location 1.
Incorrect / You marked this questionStep 1: Describe how we determine the force on a test charge at the respective locations
For a negative source charge, a positive test charge will be attracted toward the negative source charge.
Thus, you can picture the force vectors for our diagram as follows:
Per Coulomb’s Law, the force on the test charge will be equal to:
We don’t actually need to solve for the force at each point. We know that the test charge is constant at any location. Thus, the force on the test charge caused by each source charge will be proportional to the magnitude of the source charge. That is, if q_{1} is the test charge, q_{2} is the source charge.
Thus, to determine the net force at any point on the diagram, we simply add up the vectors at that location.
Step 2: Perform vector addition at each location
 Point 1: the test charge is attracted to the right with a magnitude proportional to 3e. The other test charge distances rule those forces out.
 Point 2: the test charge is attracted to the left with a magnitude proportional to 3e, and to the right with a magnitude proportional to e. Thus, the net force has a magnitude less than (proportional to) 3e
 Point 3: the test charge is attracted to the left with a magnitude proportional to e, and to the right with a magnitude proportional to 2e. Thus, the net force has a magnitude less than (proportional to) 2e
 Point 4: the test charge is to the left with a magnitude proportional to 2e. Just like for point 1, the other test charge distances rule those forces out.
As you can see, the force on the test charge is greatest at location 1.

Question 7 of 30Physics  Electrostatics, DC Circuits, Magnetism
7. Question
7. Three negative charges are located as shown in the following diagram. A charge of “e” corresponds to the charge of an electron, “2e” is twice the charge of an electron, and so on. Assume the three charges are equidistant from each other, and that each numbered location is equidistant from the charges.
We place a positive test charge into the system. If we want the test charge to move to the left, at which location should we place the charge?Correct / You marked this questionStep 1: Describe how we determine the force on a test charge at the respective locations
For a negative source charge, a positive test charge will be attracted toward the negative source charge.
Thus, you can picture the force vectors for our diagram as follows:
Per Coulomb’s Law, the force on the test charge will be equal to:
We don’t actually need to solve for the force at each point. We know that the test charge is constant at any location. Thus, the force on the test charge caused by each source charge will be proportional to the magnitude of the source charge. That is, if q_{1} is the test charge, q_{2} is the source charge.
Thus, to determine the net force at any point on the diagram, we simply add up the vectors at that location.
Step 2: Perform vector addition at each location
 Point 1: the test charge is attracted to the right with a magnitude proportional to 3e
 Point 2: the test charge is attracted to the left with a magnitude proportional to 3e, and to the right with a magnitude proportional to e. Thus, the net force on the test charge is to the left
 Point 3: the test charge is attracted to the left with a magnitude proportional to e, and to the right with a magnitude proportional to 2e. Thus, the net force on the test charge is to the right
 Point 4: the test charge is to the left with a magnitude proportional to 2e
As you can see, the test charge will be attracted towards the left at points 2 & 4
Incorrect / You marked this questionStep 1: Describe how we determine the force on a test charge at the respective locations
For a negative source charge, a positive test charge will be attracted toward the negative source charge.
Thus, you can picture the force vectors for our diagram as follows:
Per Coulomb’s Law, the force on the test charge will be equal to:
We don’t actually need to solve for the force at each point. We know that the test charge is constant at any location. Thus, the force on the test charge caused by each source charge will be proportional to the magnitude of the source charge. That is, if q_{1} is the test charge, q_{2} is the source charge.
Thus, to determine the net force at any point on the diagram, we simply add up the vectors at that location.
Step 2: Perform vector addition at each location
 Point 1: the test charge is attracted to the right with a magnitude proportional to 3e
 Point 2: the test charge is attracted to the left with a magnitude proportional to 3e, and to the right with a magnitude proportional to e. Thus, the net force on the test charge is to the left
 Point 3: the test charge is attracted to the left with a magnitude proportional to e, and to the right with a magnitude proportional to 2e. Thus, the net force on the test charge is to the right
 Point 4: the test charge is to the left with a magnitude proportional to 2e
As you can see, the test charge will be attracted towards the left at points 2 & 4

Question 8 of 30Physics  Electrostatics, DC Circuits, Magnetism
8. Question
8. We have a AA battery of 1.6 volts. How much potential energy (PE) does an electron have at the negative terminal?
The charge of an electron is 1.6 x 10^{19} Coulombs.Correct / You marked this questionStep 1: Recall how a battery operates
Chemical reactions in the battery cause electrons to accumulate at the negative terminal. The electrons then flow from the negative terminal (highest PE) to the positive terminal (lowest PE). The PE of electrons at the negative terminal is converted into other forms of energy, much in the way PE at the top of a waterfall could be converted into KE as the water descends.
Step 2: Recall our equation for electric potential energy
Step 3: Solve for electric potential energy at the negative terminal
At the negative terminal, the electric potential, V, will be at its maximum (in this case, 1.6 V)
U = qV
U = (1.6 x 10^{19} C )(1.6 V)
U = 2.56 x 10^{19} JIncorrect / You marked this questionStep 1: Recall how a battery operates
Chemical reactions in the battery cause electrons to accumulate at the negative terminal. The electrons then flow from the negative terminal (highest PE) to the positive terminal (lowest PE). The PE of electrons at the negative terminal is converted into other forms of energy, much in the way PE at the top of a waterfall could be converted into KE as the water descends.
Step 2: Recall our equation for electric potential energy
Step 3: Solve for electric potential energy at the negative terminal
At the negative terminal, the electric potential, V, will be at its maximum (in this case, 1.6 V)
U = qV
U = (1.6 x 10^{19} C )(1.6 V)
U = 2.56 x 10^{19} J 
Question 9 of 30Physics  Electrostatics, DC Circuits, Magnetism
9. Question
9. We have a AA battery of 2 volts. An electron has 1.6 x 10^{19} Joules of Potential Energy (PE). Where is the electron located within the battery?
The charge of an electron is 1.6 x 10^{19} Coulombs.Correct / You marked this questionStep 1: Recall how a battery operates
Chemical reactions in the battery cause electrons to accumulate at the negative terminal. The electrons then flow from the negative terminal (highest PE) to the positive terminal (lowest PE). The PE of electrons at the negative terminal is converted into other forms of energy, much in the way PE at the top of a waterfall could be converted into KE as the water descends.
Step 2: Recall our equation for electric potential energy
Step 3: Solve for location of the electron
U = qV
(1.6 x 10^{19} J) = (1.6 x 10^{19} C) * V
V = 1If the electric potential is 1 V where the electron is located, the electron will be in the middle of a 2V battery.
Incorrect / You marked this questionStep 1: Recall how a battery operates
Chemical reactions in the battery cause electrons to accumulate at the negative terminal. The electrons then flow from the negative terminal (highest PE) to the positive terminal (lowest PE). The PE of electrons at the negative terminal is converted into other forms of energy, much in the way PE at the top of a waterfall could be converted into KE as the water descends.
Step 2: Recall our equation for electric potential energy
Step 3: Solve for location of the electron
U = qV
(1.6 x 10^{19} J) = (1.6 x 10^{19} C) * V
V = 1If the electric potential is 1 V where the electron is located, the electron will be in the middle of a 2V battery.

Question 10 of 30Physics  Electrostatics, DC Circuits, Magnetism
10. Question
10. We have a battery of 10 volts. How much potential energy (PE) does an electron have at the positive terminal?
The charge of an electron is 1.6 x 10^{19} Coulombs.Correct / You marked this questionStep 1: Recall how a battery operates
Chemical reactions in the battery cause electrons to accumulate at the negative terminal. The electrons then flow from the negative terminal (highest PE) to the positive terminal (lowest PE). The PE of electrons at the negative terminal is converted into other forms of energy, much in the way PE at the top of a waterfall could be converted into KE as the water descends.
Step 2: Recall our equation for electric potential energy
Step 3: Solve for electric potential energy at the negative terminal
At the positive terminal, the electric potential, V, will be zero. Thus, U will also be zero.
Incorrect / You marked this questionStep 1: Recall how a battery operates
Chemical reactions in the battery cause electrons to accumulate at the negative terminal. The electrons then flow from the negative terminal (highest PE) to the positive terminal (lowest PE). The PE of electrons at the negative terminal is converted into other forms of energy, much in the way PE at the top of a waterfall could be converted into KE as the water descends.
Step 2: Recall our equation for electric potential energy
Step 3: Solve for electric potential energy at the negative terminal
At the positive terminal, the electric potential, V, will be zero. Thus, U will also be zero.

Question 11 of 30Physics  Electrostatics, DC Circuits, Magnetism
11. Question
11. What is the total resistance of the following DC circuit?
Correct / You marked this questionStep 1: First, let’s recall our equations for total resistance IN SERIES and total resistance IN PARALLEL:
Step 2: Next, let’s break down this circuit into its component parts.
Follow these order of operations: add resistors in series, then resistors in parallel.
Here are some common errors to avoid at this step:
 You may attempt to add the 1 ohm and 0.5 ohm resistors as parallel. This would be incorrect. You must first add the resistors in series before combining the resistors in parallel.
 You may think that the 2 ohm and 8 ohm resistors are in parallel. This would be incorrect. There is no line creating a fully connected loop. These resistors are actually in series, but wait until you simplify to the right before adding them.
Redrawing our correct circuit yields:
Step 3: Next, follow our “order of operations.” Add resistors in series, then add resistors in parallel. Redraw as needed.
Step 4: Take it home.
You’ve simplified it! Now all the resistors are in a series. 2 + 1 + 8 = 11 Ω
Incorrect / You marked this questionStep 1: First, let’s recall our equations for total resistance IN SERIES and total resistance IN PARALLEL:
Step 2: Next, let’s break down this circuit into its component parts.
Follow these order of operations: add resistors in series, then resistors in parallel.
Here are some common errors to avoid at this step:
 You may attempt to add the 1 ohm and 0.5 ohm resistors as parallel. This would be incorrect. You must first add the resistors in series before combining the resistors in parallel.
 You may think that the 2 ohm and 8 ohm resistors are in parallel. This would be incorrect. There is no line creating a fully connected loop. These resistors are actually in series, but wait until you simplify to the right before adding them.
Redrawing our correct circuit yields:
Step 3: Next, follow our “order of operations.” Add resistors in series, then add resistors in parallel. Redraw as needed.
Step 4: Take it home.
You’ve simplified it! Now all the resistors are in a series. 2 + 1 + 8 = 11 Ω

Question 12 of 30Physics  Electrostatics, DC Circuits, Magnetism
12. Question
12. What is the net force on charge 2 in the following diagram? Charge 1 and 2 are a electrons, Charge 3 is a proton.
The charge of an electron is 1.6 x 10^{19} Coulombs. The charge of an proton is 1.6 x 10^{19} Coulombs. The force between charge 1 & 2 is 3 Newtons. The force between charge 2 & 3 is 4 Newtons. The distance between 1 & 2 is 8.9 x 10^{18} meters. The distance between 2 & 3 is 8 x 10^{18} meters.
Correct / You marked this questionStep 1: Let’s work backwards. How do we determine the net force on an object?
This problem gives you a fair amount of extraneous information. It is important to be able to quickly parse out what is relevant for solving the problem.
The net force on charge two will simply be the vector addition of the force on charge 2 caused by charge 1, plus the force on charge 2 caused by charge 3
Step 2: Perform vector addition
We know that negative charge 1 will have a repulsive force on charge 2, and positive charge 3 will have an attractive force on charge 2. We are given the magnitudes of these forces within the problem:
From here, we simply use the Pythagorean Equation to solve for the net force (F):
Pythagorean Equation: x^{2} + y^{2} = z^{2}
(F_{x})^{2} + (F_{y})^{2} = (F)^{2}
(3 N)^{2} + (4 N)^{2} = (F)^{2}
9 N^{2} + 16 N^{2} = (F)^{2}
25 N^{2} = (F)^{2}
D = √(25 N^{2}) = 5 NIncorrect / You marked this questionStep 1: Let’s work backwards. How do we determine the net force on an object?
This problem gives you a fair amount of extraneous information. It is important to be able to quickly parse out what is relevant for solving the problem.
The net force on charge two will simply be the vector addition of the force on charge 2 caused by charge 1, plus the force on charge 2 caused by charge 3
Step 2: Perform vector addition
We know that negative charge 1 will have a repulsive force on charge 2, and positive charge 3 will have an attractive force on charge 2. We are given the magnitudes of these forces within the problem:
From here, we simply use the Pythagorean Equation to solve for the net force (F):
Pythagorean Equation: x^{2} + y^{2} = z^{2}
(F_{x})^{2} + (F_{y})^{2} = (F)^{2}
(3 N)^{2} + (4 N)^{2} = (F)^{2}
9 N^{2} + 16 N^{2} = (F)^{2}
25 N^{2} = (F)^{2}
D = √(25 N^{2}) = 5 N 
Question 13 of 30Physics  Electrostatics, DC Circuits, Magnetism
13. Question
13. Two charges are placed a certain distance from each other. Charge 1 is 1 x 10^{5} Coulombs. Charge 2 is 1 x 10^{4} Coulombs. The repulsive force between the two charges is 100 N.
How far apart are the charges?Correct / You marked this questionStep 1: Let’s recall our equation for Coulomb’s Law (i.e. for the force between two charges)
Step 2: Plug in our given values to solve for x
Incorrect / You marked this questionStep 1: Let’s recall our equation for Coulomb’s Law (i.e. for the force between two charges)
Step 2: Plug in our given values to solve for x

Question 14 of 30Physics  Electrostatics, DC Circuits, Magnetism
14. Question
14. Two equal positive charges are placed a certain distance from each other (x). What will be the relative repulsive force between the charges be if the charges are moved such that they are 4 times farther apart than the original distance?Correct / You marked this questionStep 1: Let’s recall our equation for Coulomb’s Law (i.e. for the force between two charges)
Step 2: Analyze Scenario 1 vs. Scenario 2
The distance between the charges in Scenario 2 is 4x
Scenario 1
Scenario 2
You can see the force between the two charges is ^{1}/_{16} the original force.
Incorrect / You marked this questionStep 1: Let’s recall our equation for Coulomb’s Law (i.e. for the force between two charges)
Step 2: Analyze Scenario 1 vs. Scenario 2
The distance between the charges in Scenario 2 is 4x
Scenario 1
Scenario 2
You can see the force between the two charges is ^{1}/_{16} the original force.

Question 15 of 30Physics  Electrostatics, DC Circuits, Magnetism
15. Question
15. An electric field has a strength of 1.2 x 10^{6} N/C. What is the force the electric field will have on a charge of 1 µC?Correct / You marked this questionStep 1: Let’s recall our equation for electric field strength
Step 2: Plug in our given variables to the electric field strength equation
Incorrect / You marked this questionStep 1: Let’s recall our equation for electric field strength
Step 2: Plug in our given variables to the electric field strength equation

Question 16 of 30Physics  Electrostatics, DC Circuits, Magnetism
16. Question
16. An electric field produced by a point charge has a strength of 9 x 10^{4} N/C at a certain distance from the point charge. The charge is 2.5 x 10^{4} Coulombs.
What is the distance from the point charge?Correct / You marked this questionStep 1: Let’s recall our equation for electric field produced by a point charge
Step 2: Plug in our given variables to solve for the distance from point charge (x)
Incorrect / You marked this questionStep 1: Let’s recall our equation for electric field produced by a point charge
Step 2: Plug in our given variables to solve for the distance from point charge (x)

Question 17 of 30Physics  Electrostatics, DC Circuits, Magnetism
17. Question
17. An electric field of 3 x 10^{2} N/C is produced at a certain distance from a point charge. The electric potential at this certain distance is 30 V. What is the distance from the point charge?Correct / You marked this questionStep 1: Let’s recall our equations for electric potential
We are given the electric potential and electric field strength.
Step 2: Plug in our given values to solve for distance from point charge (x)
V = Ex
30 V = (3 x 10^{2} N/C) x
x = (30 V) / (3 x 10^{2} N/C)
x = 1000 mIncorrect / You marked this questionStep 1: Let’s recall our equations for electric potential
We are given the electric potential and electric field strength.
Step 2: Plug in our given values to solve for distance from point charge (x)
V = Ex
30 V = (3 x 10^{2} N/C) x
x = (30 V) / (3 x 10^{2} N/C)
x = 1000 m 
Question 18 of 30Physics  Electrostatics, DC Circuits, Magnetism
18. Question
18. What is the standard unit of magnetic field?Correct / You marked this questionLet’s consider each option:
[A] Farad
Farad is the standard unit of capacitance (storage potential for a capacitor)
[B] Joule
Joule is the unit of energy
[C] Tesla
This is the correct answer.
[D] Volt
Volts are the standard unit of electric potential between two points. “Voltage” is the electric potential energy per unit charge, i.e joules / coulomb.
[E] Coulomb
Coulombs are the standard unit of charge.
Incorrect / You marked this questionLet’s consider each option:
[A] Farad
Farad is the standard unit of capacitance (storage potential for a capacitor)
[B] Joule
Joule is the unit of energy
[C] Tesla
This is the correct answer.
[D] Volt
Volts are the standard unit of electric potential between two points. “Voltage” is the electric potential energy per unit charge, i.e joules / coulomb.
[E] Coulomb
Coulombs are the standard unit of charge.

Question 19 of 30Physics  Electrostatics, DC Circuits, Magnetism
19. Question
19. A charge is stationary in a magnetic field. The charge is 12 Coulombs. The strength of the magnetic field is 4 Teslas.
What is the force on the charge from the magnetic field?Correct / You marked this questionStep 1: Let’s recall our equation for magnetic force
Step 2: Analyze the magnetic force equation with our given variables
We are told that the charge is stationary. Thus, the charge has zero velocity, and thus, experiences a force of 0 N. Only moving charges through a magnetic field experience a magnetic force.
Incorrect / You marked this questionStep 1: Let’s recall our equation for magnetic force
Step 2: Analyze the magnetic force equation with our given variables
We are told that the charge is stationary. Thus, the charge has zero velocity, and thus, experiences a force of 0 N. Only moving charges through a magnetic field experience a magnetic force.

Question 20 of 30Physics  Electrostatics, DC Circuits, Magnetism
20. Question
20. A charge is moving through a magnetic field parallel to the direction of the magnetic field. The charge is 5 Coulombs and moving with a velocity of 5 m/s. The strength of the magnetic field is 5 Teslas.
What is the force on the charge from the magnetic field?Correct / You marked this questionStep 1: Let’s recall our equation for magnetic force
Step 2: Analyze the magnetic force equation with our given variables
We are told that the charge is moving in a direction parallel to the magnetic field. Thus, θ = 0, and sinθ = 0.
As such, the force (F) = 0 N.
Incorrect / You marked this questionStep 1: Let’s recall our equation for magnetic force
Step 2: Analyze the magnetic force equation with our given variables
We are told that the charge is moving in a direction parallel to the magnetic field. Thus, θ = 0, and sinθ = 0.
As such, the force (F) = 0 N.

Question 21 of 30Physics  Electrostatics, DC Circuits, Magnetism
21. Question
21. A positive charge is moving through a magnetic field. The charge is 12 Coulombs and moving with a velocity of 8 m/s. The strength of the magnetic field is 7 Teslas.
 The magnetic force on the charge is in the direction XXX (into the page)
 The velocity of the charge is in the direction of directly upward vertical (perpendicular to magnetic force).
What is the direction of the magnetic field (B)?Correct / You marked this questionStep 1: Recall our “Right Hand Rule” conventions for a charge moving through a magnetic field
Step 2: Deduce the direction of the magnetic field (B) using the right hand rule
We are told that the charge is positive.
 Point your thumb in the direction of F (into the page)
 Point your index finger in the direction of v (upward vertical, perpendicular to F)
 Point your middle finger in the direction of B
Our middle finger will point towards the right →
Incorrect / You marked this questionStep 1: Recall our “Right Hand Rule” conventions for a charge moving through a magnetic field
Step 2: Deduce the direction of the magnetic field (B) using the right hand rule
We are told that the charge is positive.
 Point your thumb in the direction of F (into the page)
 Point your index finger in the direction of v (upward vertical, perpendicular to F)
 Point your middle finger in the direction of B
Our middle finger will point towards the right →

Question 22 of 30Physics  Electrostatics, DC Circuits, Magnetism
22. Question
22. A negative charge is moving through a magnetic field. The charge is 86 Coulombs and moving with a velocity of 180 m/s. The strength of the magnetic field is 100 Teslas.
 The magnetic force on the charge is in the direction to the left (←)
 The magnetic field is in the direction directly upward vertical (perpendicular to magnetic force).
What is the direction of the velocity of the charge?Correct / You marked this questionStep 1: Recall our “Right Hand Rule” conventions for a charge moving through a magnetic field
Step 2: Deduce the direction of the velocity of the charge (v) using the right hand rule
We are told that the charge is negative.
 Point your thumb in the direction OPPOSITE of F (to the right)
 Point your middle finger in the direction of B (directly upward vertical (perpendicular to magnetic force)
 Point your index finger in the direction of v
Our index finger will point into the page (XXX)
Incorrect / You marked this questionStep 1: Recall our “Right Hand Rule” conventions for a charge moving through a magnetic field
Step 2: Deduce the direction of the velocity of the charge (v) using the right hand rule
We are told that the charge is negative.
 Point your thumb in the direction OPPOSITE of F (to the right)
 Point your middle finger in the direction of B (directly upward vertical (perpendicular to magnetic force)
 Point your index finger in the direction of v
Our index finger will point into the page (XXX)

Question 23 of 30Physics  Electrostatics, DC Circuits, Magnetism
23. Question
23. A wire is carrying a current through a magnetic field as shown below. The magnetic field is coming out of the page
(◆ denotes “out of the page” i.e coming towards you)
What is the direction of the magnetic force on the wire?Correct / You marked this questionStep 1: Recall our “Right Hand Rule” conventions for a current carrying wire
Step 2: Deduce the direction of the magnetic force on the wire (F) using the right hand rule
We are deducing the magnetic force ON the wire; thus, we will use the “determine direction of F” box above.
 Point your index finger in the direction of I (to the right). Do not overthink the charge of the current, it will be positive current/charges unless stated otherwise.
 Point your middle finger in the direction of B (◆◆◆ out of page)
 Point your thumb in the direction of F
Your thumb will point down (↓)
Incorrect / You marked this questionStep 1: Recall our “Right Hand Rule” conventions for a current carrying wire
Step 2: Deduce the direction of the magnetic force on the wire (F) using the right hand rule
We are deducing the magnetic force ON the wire; thus, we will use the “determine direction of F” box above.
 Point your index finger in the direction of I (to the right). Do not overthink the charge of the current, it will be positive current/charges unless stated otherwise.
 Point your middle finger in the direction of B (◆◆◆ out of page)
 Point your thumb in the direction of F
Your thumb will point down (↓)

Question 24 of 30Physics  Electrostatics, DC Circuits, Magnetism
24. Question
24. What is the standard unit of current?Correct / You marked this questionLet’s consider each option:
[A] Farad
Farad is the standard unit of capacitance (storage potential for a capacitor)
[B] Ampere
This is the correct answer
[C] Tesla
Tesla is the standard unit of magnetic field
[D] Volt
Volts are the standard unit of electric potential between two points. “Voltage” is the electric potential energy per unit charge,
i.e joules / coulomb.[E] Coulomb
Coulombs are the standard unit of charge.
Incorrect / You marked this questionLet’s consider each option:
[A] Farad
Farad is the standard unit of capacitance (storage potential for a capacitor)
[B] Ampere
This is the correct answer
[C] Tesla
Tesla is the standard unit of magnetic field
[D] Volt
Volts are the standard unit of electric potential between two points. “Voltage” is the electric potential energy per unit charge,
i.e joules / coulomb.[E] Coulomb
Coulombs are the standard unit of charge.

Question 25 of 30Physics  Electrostatics, DC Circuits, Magnetism
25. Question
25. The currentcarrying wire below is generating a magnetic field.
Which of the following best describes the direction of the magnetic field?
Note:
(◆ denotes “out of the page” i.e coming towards you)
(X denotes “into the page” )Correct / You marked this questionStep 1: Recall our “Right Hand Rule” conventions for a current carrying wire
Step 2: Deduce the direction of the magnetic field (B) using the right hand rule
We are asked to analyze the magnetic field (B) GENERATED BY the wire. Thus we will use the “Generated By Current” box above.
 Thumb points in direction of I
 Fingers curl in direction of B
We can see that this aligns with
Incorrect / You marked this questionStep 1: Recall our “Right Hand Rule” conventions for a current carrying wire
Step 2: Deduce the direction of the magnetic field (B) using the right hand rule
We are asked to analyze the magnetic field (B) GENERATED BY the wire. Thus we will use the “Generated By Current” box above.
 Thumb points in direction of I
 Fingers curl in direction of B
We can see that this aligns with

Question 26 of 30Physics  Electrostatics, DC Circuits, Magnetism
26. Question
26. The currentcarrying wire below is generating a magnetic field.
Which of the following best describes the direction of the magnetic field generated by the wire?
Note:
(◆ denotes “out of the page” i.e coming towards you)
(X denotes “into the page” )Correct / You marked this questionStep 1: Recall our “Right Hand Rule” conventions for a current carrying wire
Step 2: Deduce the direction of the magnetic field (B) using the right hand rule
We are asked to analyze the magnetic field (B) GENERATED BY the wire. Thus we will use the “Generated By Current” box above.
 Thumb points in direction of I
 Fingers curl in direction of B
We can see that inside the wire, the magnetic field will point out of the page, while outside of the wire, the magnetic field will point into the page.Thus, the direction of the magnetic field depends on the location relative to the wire.
Incorrect / You marked this questionStep 1: Recall our “Right Hand Rule” conventions for a current carrying wire
Step 2: Deduce the direction of the magnetic field (B) using the right hand rule
We are asked to analyze the magnetic field (B) GENERATED BY the wire. Thus we will use the “Generated By Current” box above.
 Thumb points in direction of I
 Fingers curl in direction of B
We can see that inside the wire, the magnetic field will point out of the page, while outside of the wire, the magnetic field will point into the page.Thus, the direction of the magnetic field depends on the location relative to the wire.

Question 27 of 30Physics  Electrostatics, DC Circuits, Magnetism
27. Question
27. An Amp is equivalent to:Correct / You marked this questionThis is a conceptual question.
Amps are the standard unit of electric current. Amps are a measure of the rate of electric charge passing through a point in a circuit.
If we know that amps are a measure of rate, we can deduce that the denominator in our answer options should be a unit of time. We know that electric charge (our numerator) is measured in coulombs.
You can use a waterflow pipe analogy help understand amps / current: think of current, (i.e. amperage) as the rate of water flowing through a pipe.
Incorrect / You marked this questionThis is a conceptual question.
Amps are the standard unit of electric current. Amps are a measure of the rate of electric charge passing through a point in a circuit.
If we know that amps are a measure of rate, we can deduce that the denominator in our answer options should be a unit of time. We know that electric charge (our numerator) is measured in coulombs.
You can use a waterflow pipe analogy help understand amps / current: think of current, (i.e. amperage) as the rate of water flowing through a pipe.

Question 28 of 30Physics  Electrostatics, DC Circuits, Magnetism
28. Question
28. The current in a DC circuit is 10 Amps. How many coulombs of charge pass through the current in 5 seconds?Correct / You marked this questionStep 1: Recall our Current Equation for a DC Circuit
Step 2: Plug in our given variables to solve for charge
I = q/t
10 A = q / (5 seconds)
q = 50 A*seconds = 50 CoulombsIncorrect / You marked this questionStep 1: Recall our Current Equation for a DC Circuit
Step 2: Plug in our given variables to solve for charge
I = q/t
10 A = q / (5 seconds)
q = 50 A*seconds = 50 Coulombs 
Question 29 of 30Physics  Electrostatics, DC Circuits, Magnetism
29. Question
29. What is the standard unit of resistance?Correct / You marked this questionThis is a conceptual question.
Ohms are the standard unit of resistance in a circuit. Resistance is measured between two points in a circuit. You can use a waterflow pipe analogy help understand resistance: think of resistance as the “friction” slowing down the moving water.
Incorrect / You marked this questionThis is a conceptual question.
Ohms are the standard unit of resistance in a circuit. Resistance is measured between two points in a circuit. You can use a waterflow pipe analogy help understand resistance: think of resistance as the “friction” slowing down the moving water.

Question 30 of 30Physics  Electrostatics, DC Circuits, Magnetism
30. Question
30. In a DC circuit, current:Correct / You marked this questionLet’s analyze each answer option individually:
[A] moves in the direction of negative charge flow
This is incorrect. In a DC Circuit, current moves in the direction of positive charge flow.
[B] moves in the direction of positive charge flow
This is the correct answer. This is actually an arbitrary choice that has stuck as convention.
[C] is equal to R/V
Current is actually equal to V/R:
[D] is in units of amps / second
Current is measured in units of amps. Amps are equivalent to coulombs / second.
[E] is typically greater with increased temperature
In most cases, an increase temperature increases the resistance in a DC circuit, thus decreasing the current.
STRATEGY TIP:
Answer options A & B are directly opposing: i.e. if one is true, the other one must be false. Thus, we can deduce the answer is very likely to be one of these options.Incorrect / You marked this questionLet’s analyze each answer option individually:
[A] moves in the direction of negative charge flow
This is incorrect. In a DC Circuit, current moves in the direction of positive charge flow.
[B] moves in the direction of positive charge flow
This is the correct answer. This is actually an arbitrary choice that has stuck as convention.
[C] is equal to R/V
Current is actually equal to V/R:
[D] is in units of amps / second
Current is measured in units of amps. Amps are equivalent to coulombs / second.
[E] is typically greater with increased temperature
In most cases, an increase temperature increases the resistance in a DC circuit, thus decreasing the current.
STRATEGY TIP:
Answer options A & B are directly opposing: i.e. if one is true, the other one must be false. Thus, we can deduce the answer is very likely to be one of these options.
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