15.4. Magnetism Problem Set http://www.ck12.org
e. Show the path and calculate the radius of the positron in region II.
f. Now the magnetic field is removed; calculate the acceleration of the positron away from the center.
g. Calculate the angle away from the center with which it would enter region II if the magnetic field were
to be removed.
- A small rectangular loop of wire 2.00 m by 3.00 m moves with a velocity of 80.0 m/s in a non-uniform field
that diminishes in the direction of motion uniformly by.0400 T/m. Calculate the induced emf in the loop.
What would be the direction of current? - An electron is accelerated through 20,000 V and moves along the positivex−axis through a plate 1.00 cm
wide and 2.00 cm long. A magnetic field of 0.020 T is applied in the−zdirection.
a. Calculate the velocity with which the electron enters the plate.
b. Calculate the magnitude and direction of the magnetic force on the electron.
c. Calculate the acceleration of the electron.
d. Calculate the deviation in theydirection of the electron form the center.
e. Calculate the electric field necessary to keep the electron on a straight path.
f. Calculate the necessary voltage that must be applied to the plate. - A long straight wire is on thex−axis and has a current of 12 A in the−xdirection. A pointP, is located 2.0 m
above the wire on they−axis.
a. What is the magnitude and direction of the magnetic field atP.
b. If an electron moves throughPin the−xdirection at a speed of 8. 0 × 107 m/s what is the magnitude and
direction of the force on the electron?
c. What would be the magnitude and direction of an electric field to be applied atPthat would counteract
the magnetic force on the electron?
Answers to Selected Problems
- No: ifv=0 thenF=0; yes:F=qE
2..
3.. - a. Into the page b. Down the page c. Right
- Both pointing away from north
6..
7.. - 7.6 T, south
- Down the page; 60 N
- a. To the right, 1. 88 × 104 N b. 91.7 m/s c. It should be doubled
- East 1. 5 × 104 A
- 0.00016 T; if CCW motion, B is pointed into the ground.
- 2 ×105 V, counterclockwise
- a. 15 V b. Counter-clockwise
- a. 2× 10 −^5 T b. Into the page c. 2.8 N/m d. CW
- a. 2. 42 × 108 m/s b. 9. 69 × 10 −^12 N c..0055 m
- E/B
- a. 8× 10 −^7 T b. 1. 3 × 10 −^6 C
- a. 0.8 V b. CCW c..064 N d..16 N/C e..13 w
- a. 1. 11 × 108 m/s b. 9. 1 × 10 −^30 N<< 6. 4 × 10 −^14 N d..00364 T e..173 m f. 7. 03 ×1016 m/s^2 g. 3. 27 ◦
- 19.2 V
- a.8. 39 × 107 m/s b. 2. 68 × 10 −^13 N,−yc. 2. 95 × 101 7 m/s^2 d..00838 m e. 1. 68 × 106 N/C f. 16,800 V
- a. 1. 2 × 10 −^6 T,+zb. 1. 5 × 10 −^17 N,−yc. 96 N/C,−y