What if the initial velocity of the charged particle is not necessarily perpendicular to the magnetic field
B? In that case, the motion will have two components: motion in a circle perpendicular to the magnetic field,
combined with uniform motion parallel to the magnetic field. The net motion will be that of ahelix: the
particles will spiral around magnetic field lines in helices.
How big are the circles that a charged particle moves in? We can find that by equating the magnetic force
of Eq. (32.1) (withv?B) to the centripetal forcemv^2 =r. We then haveqvBDmv^2 =r; solving for the
radiusrof the circle, we haverDmv=.qB/. More generally, if the particle is moving in a helix, then the
radius of the helix is determined by the component of the particle’s velocityvthat is perpendicular to the
magnetic field (v?). Also, since the radius is always positive, we want to use the absolute value of the charge
q. The general result is that the radius of the helix is
rDmv?
jqjB: (32.3)
This radius is called thegyroradius,cyclotron radius,orLarmor radius. The gyroradius will be larger for a
weaker magnetic field, or for a heavier or faster particle.
Another important quantity is the angular frequency with which the particle gyrates in a circle about the
magnetic field lines. The time it takes the particle to complete one circle (i.e. the period of the motion) is the
total distance divided by the speed:TD2r=v?. Substitutingrfrom Eq. (32.3), we haveTD2m=.jqjB/.
Since the angular frequency!D2=T, we have that angular frequency of the motion as
!D
jqjB
m: (32.4)
This is called thegyrofrequencyorcyclotron frequency. A particle will spin around in circles faster for a
stronger magnetic field or a lighter particle.
32.2 Force on a Wire in a Magnetic Field
Now suppose we have a wire carrying in an electric currentIplaced within a magnetic fieldB. Within the
wire, the current is being carried by electrons moving with the drift velocity, each of which experiences a
Lorentz force. There will then be a forceFon the wire given by
FDIlB: (32.5)HereIis the current, andlis a vector whose length is equal to the length of the wire and which points in the
direction of the conventional current. Applying this to a current loop, for example, gives the same torque as
given by Eq. (31.11).
32.3 Magnetic Force between Two Long Wires
If we put two long wires next to each other so that they are parallel, then each wire generates a magnetic field
that envelopes the other wire. By combining Eq. (31.8) (which gives the magnetic field generated by a wire)
with Eq. (32.5) (which gives the force on a wire in a magnetic field), we can find the mutual force between
the two parallel wires. The result is
F
`D
0
2
I 1 I 2
d; (32.6)
whereF=`is the force per unit length,I 1 andI 2 are the two currents, anddis the distance between the two
wires.