Transcript
47
Marks Total
ASSIGNMENT 8
Part One: Moving Charges in Magnetic Fields
Part One of this assignment is worth 17 marks. The value of each question is noted in parentheses in the left margin. Note: The answer areas will expand to fit the length of your response.
1.
State the direction of the magnetic force on the moving charge in each diagram below. You may wish to verify your answers using the online simulation.
(1)
positive charge
Answer:
Up
(1)
positive charge
Answer:
Down
(1)
negative charge
Answer:
Up
(1)
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negative charge
Answer:
Down
(4) 2.
An electron experiences a downward magnetic force of 7.20 × 10-14 N when it is travelling at 3.00 × 105 m/s south through a magnetic field. Calculate the magnitude of the magnetic field and determine its direction using the left-hand rule. Explain all finger directions and the palm direction.
Answer:
B= F/qv
= (7.20*10-14) /(1.6*10-19 * 3*10-5) = 1.5T (tesla)
The thumb is force, index is vector of velocity, middle is the magnetic field.
(2) 3.
A charged particle is travelling west through a downward magnetic field and it experiences a magnetic force directed to the north. Using the appropriate hand rule, determine if the charge is negative or positive. Explain all finger directions and the palm direction.
Answer:
Negative
(4) 4.
Calculate the magnitude and the direction of the magnetic force acting on an alpha particle that is travelling upwards at a speed of 3.00 x 105 m/s through a 0.525 T west magnetic field. Explain all finger directions and the palm direction.
Answer:
(3) 5.
An electron (m = 9.11 × 10-31 kg) enters a downward magnetic field of 5.00 × 10-1 T with a velocity of 6.50 × 106 m/s West. Calculate the radius of the circular path it will follow once it is travelling within the magnetic field.
Answer:
R = mv /bq
(9.11 x 10^-31 kg)(6.50 x 10^6 m/s) /(1.60 x 10^-19 C)(5.00 x 10^-1 T)
= 7.40 x 10-5
STOP!
When you have completed all of the questions in Part One, save your work to your desktop. You will return to this assignment to complete Part Two after you have completed the remainder of the content in the next section.
Part Two: Electromagnetic Induction
Part Two of this assignment is worth 30 marks. The value of each question is noted in the left margin in parenthesis. Note: The answer areas will expand to fit the length of your response.
1.
(2)
A compact fluorescent light bulb draws a current of 0.10 A for one hour.
How much charge flows through the bulb in one hour?
Answer:
Current I = q / t
q = I * t = 0.10A * 3600s = 360
(2)
How many electrons flow through the light bulb in the hour? Remember the charge of one electron is the elementary charge on your physics data sheet.
Answer:
360 C / 1.6*10^-19 C
= 2.25*10^21
(2) 2.
A motor uses a coil of wire in a magnetic field to generate force. The motor draws a current of 9.50 A through the coil of wire and has a magnetic field of 1.75 T. If the motor is designed to generate 800 N, how long is the wire in the coil assuming that all of the wire creates force?
Answer:
B = 0.525 Tq = 3.20 x 10^-19 Cv = 3.00 x 10^5 m/sF = qvbF = (3.20 x 10^-19 C)(3.00 x 10^5 m/s)(0.525 T
(2) 3.
What would happen to the direction of the magnetic force if the loop of wire were to undergo one half turn without reversing the direction of the current in the loop?
Answer:
the current is moving clockwise, and the field is left to right. Using the left hand rule, the left hand side of the loop would be moving down, and the right hand side up. If the loop is now turned over by 180 degrees with reversing the current, the left hand side of the loop will now move up. If the loop is free, it will oscillate to and fro and eventually settle so that it is vertical and at right angles to the magnetic field of the magnet. Its the job of the commutator to effectively reverse the direction of the current, every half cycle, to ensure that the loop continues to turn in one direction.
4.
In the diagrams below, a magnet is either dropped down or pulled up through a cylinder encircled by a coil conductor. Depending on what is asked for in questions a through e, (the “?”), indicate whether the magnetic pole is north or south and/or whether the motion of the magnet is up or down. Remember that current is e- flow.
(1)
Determine the magnetic poles of the cylinder.
Answer:
South (top), North (bottom)
(1)
Determine the magnetic poles of the cylinder.
Answer:
North (top), South (bottom)
(1)
Determine if the magnet is pulled up through the cylinder or pushed down through the cylinder.
Answer:
pulled up through the cylinder
(1)
Determine the magnetic poles of the cylinder and the magnet.
Answer:
(1)
Determine if the magnet is pulled up through the cylinder or pushed down through the cylinder and determine the magnetic poles of the cylinder.
Answer:
(1)
Determine if the magnet is pulled up through the cylinder or pushed down through the cylinder.
Answer:
pulled up through
5.
(2)
Knowing the formulas for magnetic force, describe how each of the following factors influences the magnitude of the magnetic force and give the reason.
Strength of the magnetic field
Answer:
F = qvB, if B increases so does F (proportional)
(2)
Speed of the magnet passing through the coil
Answer:
F = qvB, if v increases so does F (proportional)
(2)
The number of loops wrapped around a cylinder
Answer:
F = nBIL, if n increases so does F (proportional)
(2)
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The current
Answer:
F = nBIL, if I increases, so does F (proportional)
(2) 6.
Explain why the generator effect would occur in the same way if the wire was moved rather than the magnet.
Answer:
Generator effect occurs when there is a change in magnetic field when the flux lines cut each other producing voltage EMF=vlB, hence does not matter if the magnet or the wire moves, cause either way, their movement will cause a change in flux hence creating voltage which allows current to flow; the generator effect.
(2) 7.
Would two parallel wires carrying a current in opposite directions repel or attract one another? Explain your answer.
Answer:
It will Repel. If the current is in the opposite direction, again turn your hand such that the index finger faces towards the current. You will see force is away. If one is north and one is south, their field lines, focusing at the middle of the two wires, would be disrupting and unable flow in same direction of the ends of the wire use the right hand solenoid hand curl rule) so the lines cannot connect
(4) 8.
A 0.120 m long copper wire has a mass of 9.02 g and is carrying a current of 5.10 A perpendicular to a uniform magnetic field. The apparatus is placed in a strong magnetic field and the wire is found to levitate. Calculate the magnetic field strength. Remember to show all work.
Answer:
F = BIL
F = ma
ma = Bil
MA /Il = B
= (.00902)(9.8) /(.120)(5.10) = B
= B = 0.14 Tesla
When you have completed all of the questions in this assignment,
submit your work to your teacher
