Introduction to SAT II Physics - FreeExamPapers

A particle with a positive charge of 3 C moves upward at a speed of 10 m/s. It passes
simultaneously through a magnetic field of 0.2 T directed into the page and an electric field
of 2 N/C directed to the right. How is the motion of the particle affected?
Answering this question is a matter of calculating the force exerted by the magnetic field
and the force exerted by the electric field, and then adding them together. The force
exerted by the magnetic field is:
Using the right-hand-rule, we find that this force is directed to the left. The force exerted
by the electric field is:
This force is directed to the right. In sum, we have one force of 6 N pushing the particle to
the left and one force of 6 N pushing the particle to the right. The net force on the particle
is zero, so it continues toward the top of the page with a constant velocity of 10 m/s.
Magnetic Force on Current-Carrying Wires
Since an electric current is just a bunch of moving charges, wires carrying current will be
subject to a force when in a magnetic field. When dealing with a current in a wire, we
obviously can’t use units of q and v. However, qv can equally be expressed in terms of Il,
where I is the current in a wire, and l is the length, in meters, of the wire—both qv and Il
are expressed in units of C · m/s. So we can reformulate the equation for the magnitude of
a magnetic force in order to apply it to a current-carrying wire:
In this formulation, is the angle the wire makes with the magnetic field. We determine
the direction of the force by using the right-hand rule between the direction of the current
and that of the magnetic field.
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