# 47 Marks Total ASSIGNMENT 8   Part One:  Moving Charges in Magnetic Fields   Part One of this assignment is worth 17 marks.  The value of...

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 47Marks 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) a.    positive charge Answer: (1) b.    positive charge Answer: (1) c.     negative charge Answer: (1) d.    negative charge Answer: (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)  Use flemings left hand motor rule. Where thumb is force, index is field, middle is current. Its difficult to say without seeing a diagram of the problem it will be at 90 degrees though. You can work it out from the motor rule though (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: Electron, as only the left hand palm rule worked, if you tried the right hand palm rule it doesn't work out and right hand palm rule with the thumb representing current, it represents direction of velocity in this case and right hand represents the movement of protons (i like to think of it as, right hand, is my pro hand hence proton) therefore left hand must be electron, which is how i arrived at my answer. Another alternative is the flemming's rule, same principle as what i said about, thumb points at force, index finger at magnetic field and middle finger at current (velocity in this case).

 (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.

 (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 current is in opposite direction, again turn your hand such that second 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=BILF=mama=BilMA/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

• Posted: 3 years ago
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