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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. |
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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. |
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Solve for B |
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The direction of the force is
the result of the right hand rule. In
the figure below which is drawn from above looking down, |
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, so putting fingers along (to the right as drawn) and curing fingers through the smallest angle for up (out of page as drawn) your right thumb will be pointing towards bottom of page which is south as drawn. So we know must be out of page or up |
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In the figure below repeat
process but this time we are looking horizontally and we still see to get a
force pointing south, must point up! |
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For electron |
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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 |
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