|
|||||||||||||||||||
2. (a) A proton, traveling with a velocity of 4.5 x 106 m/s due east, experiences a magnetic force that has a
maximum magnitude of 8.0 x 10-14 N and a direction of due south. What are the
magnitude and direction of the magnetic field causing the force? (b) Repeat part (a) assuming the
proton is replaced by an electron. |
|||||||||||||||||||
|
|||||||||||||||||||
Since the maximum magnitude of
force is found, assume the B field must be perpendicular to velocity as sin(q) is a maximum
then. So the magnitude of the field can be found from the magnetic force equation. |
|||||||||||||||||||
|
|||||||||||||||||||
|
|||||||||||||||||||
Solve for B |
|||||||||||||||||||
|
|||||||||||||||||||
|
|||||||||||||||||||
The direction of the force is
the result of the right hand rule. In
the figure below which is drawn from above looking down, |
|||||||||||||||||||
|
|||||||||||||||||||
|
|||||||||||||||||||
In the figure below repeat
process but this time we are looking horizontally and we still see to get a
force pointing south, |
|||||||||||||||||||
|
|||||||||||||||||||
|
|||||||||||||||||||
For electron |
|||||||||||||||||||
The only change is the charge
is now –e instead of +e so the minus sign changes the direction from up to
down to continue to have force in the south direction. The magnitude remains 0.111 T |
|||||||||||||||||||
|
|||||||||||||||||||
|
|||||||||||||||||||
|
|||||||||||||||||||
|
|||||||||||||||||||
|
|||||||||||||||||||
|