**Problem : **

A uniform magnetic field in the positive *y* direction acts on a positively
charged particle moving in the positive *x* direction. In what direction does the
force act on particle?

To solve this problem we simply use the right hand rule. First we construct a
three dimensional axis, as shown below. Then we point our thumb in the positive
*x* direction, our index finger in the positive *y* direction, and we find that our
middle finger points in the positive *z* direction, implying that this is exactly
the direction of the force on the particle.

The magnetic field, direction of charge motion, and resultant force shown inr
relation to a coordinate axis

**Problem : **

Two vectors, *v*_{1} and *v*_{2}, each with magnitude of 10, act in the *x*-*y* plane,
at an angle of 30^{o}, as shown below. What is the magnitude and
direction of the cross product *v*_{1}×*v*_{2}?

Two vectors in the *x*-*y* plane. What is their cross product?

Finding the magnitude of the cross product is easy: it is simply *v*_{1}*v*_{2}sin*θ* = (10)(10)(.5) = 50. The direction of the cross product, however, takes a
little thought. Since we are computing *v*_{1}×*v*_{2}, think of *v*_{1} as a
velocity vector, and *v*_{2} as a magnetic field vector. Using the right hand
rule, then, we find that the cross product of the two points in the positive *z*
direction. Notice from this problem that cross products are not communative:
the direction of *v*_{1}×*v*_{2} is the opposite of that of *v*_{2}×*v*_{1}.
This problem should help with the complicated directions of fields, velocities,
and forces.

**Problem : **

A uniform electric field of 10 dynes/esu acts in the positive *x* direction, while
a uniform magnetic field of 20 gauss acts in the positive *y* direction. A
particle of charge *q* and velocity of .5*c* moves in the positive *z* direction.
What is the net force on the particle?

To solve the problem we use the equation:

= q + |

So we must find the vector sum of the electric force and the magnetic force. The electric force is easy: it is simply

**Problem : **

A charged particle moving perpendicular to a uniform magnetic field always experiences a net force perpendicular to its motion, similar to the kind of force experienced by particles moving in uniform circular motion. The magnetic field can actually cause the particle to move in a complete circle. Express the radius of this circle in terms of the charge, mass and velocity of the particle, and the magnitude of the magnetic field.

In this case the magnetic field produces the centripetal
force required to move the
particle in uniform circular motion. We know that, since *v* is perpendicular to
*B*, the magnitude of the magnetic force is simply *F*_{B} = . We also
know that any centripetal force has magnitude *F*_{c} = . Since the
magnetic force is the only one acting in this situation, we may relate the two
quantities:

F_{c} | = | F_{B} | |

= | |||

mv^{2}c | = | qvBr | |

r | = |

Analyzing our answer we see that stronger fields cause particles to move in smaller circles.

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