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.
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} ?
Finding the magnitude of the cross product is easy: it is simply v _{1} v _{2}sinθ = (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 .5c moves in the positive z direction. What is the net force on the particle?
To solve the problem we use the equation:
= q + |
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 | = |