Peoples Physics Book Version-2

15.3. Effects of Magnetic Fields http://www.ck12.org

Since the force on a charged particle in a magnetic field is always perpendicular to both its velocity vector and the
field vector (check this using the second right hand rule above), aconstantmagnetic field will provide a centripetal
force — that is, a constant force that is always directed perpendicular to the direction of motion. Two such
force/velocity combinations are illustrated above. According to our study of rotational motion, this implies that
as long as the particle does not leave the region of the magnetic field, it will travel in a circle. To find the radius of
the circle, we set the magnitude of the centripetal force equal to the magnitude of the magnetic force and solve forr:

Fc=

mv^2

r

=FB=qvBsinθ=qvB

Therefore,

r=

mv^2

qvB

In the examples above,θwas conveniently 90 degrees, which made sinθ=1. But that does not really matter; in
a constant magnetic fields a differentθwill simply decrease the force by a constant factor and will not change the
qualitative behavior of the particle, sinceθcannot change due to such a magnetic force.(Why? Hint: what is the
force perpendicular to? Read the paragraph above.)

Force on a Wire

Since a wire is nothing but a collection of moving charges, the force it will experience in a magnetic field will simply
be the vector sum of the forces on the individual charges. If the wire is straight — that is, all the charges are moving
in the same direction — these forces will all point in the same direction, and so will their sum. Then, the direction
of the force can be found using the second right hand rule, while its magnitude will depend on the length of the wire
(denotedL), the strength of the current, the strength of the field, and the angle between their directions:

Fwire=LIBsin(θ) [4] Force on a Current Carrying Wire

Two current-carrying wires next to each other each generate magnetic fields and therefore exert forces on each other: