Peoples Physics Concepts

14.2. Lorentz Force http://www.ck12.org

Now, plugging the known values we have:

B=

FB

qvsinθ

=

2 .2? 10−^12 N

1. 6 × 10 −^19 C× 1. 8 × 106 m/s× 1

= 7 .6 T

Now we will find the direction of the field. We know the direction of the velocity (east) and the direction of the

force due to the magnetic field (up, out of the page). Therefore we can use the second right hand rule (we will use

the left hand, since an electron’s charge is negative). Point the pointer finger to the right to represent the velocity

and the thumb up to represent the force. This forces the middle finger, which represents the direction of the

magnetic field, to point south. Alternatively, we could recognize that this situation is illustrated for apositive

particle in the right half of the drawing above; for a negative particle to experience the same force, the field has to

point in the opposite direction: south.

Example 2: Circular Motion in Magnetic Fields

Consider the following problem: a positively charged particle with an initial velocity of~v 1 , chargeqand massm

traveling in the plane of this page enters a region with a constant magnetic field~Bpointing into the page. We are

interested in finding the trajectory of this particle.

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.