Introductory Physics Volume Two
Introductory Physics Volume Two
Introductory Physics Volume Two
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84 Magnetic Fields 4.3<br />
For a magnetic force, ⃗v and ⃗ F are prependicular so that ⃗ F ·⃗v = 0. Thus<br />
dW = 0 for a magnetic force.<br />
Theorem: Work by Magnetic Force<br />
The magnetic force does zero work. Thus the magnetic force can<br />
change the direction but not the speed of a particle.<br />
Because the magnetic force changes only the direction of a particle,<br />
a magnetic field is useful for steering charged particles, once they are<br />
already moving.<br />
Consider a particle that is in a region with a uniform magnetic<br />
field. Suppose that the particles initial velocity is perpendicular to the<br />
magnetic field. Since the magnetic force is always perpendicular to<br />
the direction of the magnetic field, the force will have no component<br />
parallel to the field. So, since the particle started with zero velocity<br />
parallel to the field it will continue to have zero velocity parallel to the<br />
field. But this means that the velocity will always be perpendicular to<br />
the magnetic field, so |⃗v × B| ⃗ = vB. Thus we know that the magnitude<br />
of the magnetic force is<br />
F ≡ | ⃗ F | = q|⃗v × ⃗ B| = qvB<br />
In addition we know by the previous theorem that the speed of the<br />
particle will be constant, v = v 0 . So that the magnitude of the force<br />
is constant. In summary, we see that the magnitude of the magnetic<br />
force on the particle is constant and perpendicular to the velocity, so<br />
that the acceleration of the particle is constant and perpendicular to<br />
the velocity. We have run into this situation before: In circular motion<br />
the acceleration is constant and always perpendicular to the velocity.<br />
We are lead by this observation to the following theorem.<br />
Theorem: Circular Trajectories<br />
If a particle is in a region of uniform magnetic field and entered<br />
the region with a velocity perpendicular to the magnetic field, then<br />
the particle will execute circular motion while in the region.<br />
One can relate the radius and velocity of the motion to the field<br />
strength and charge by employing Newton’s second law.<br />
F = ma<br />
−→ qvB = m v2<br />
r<br />
⊲ Problem 4.3<br />
One can use the circular motion of a charge particle to determine it’s<br />
mass if you already know it’s charge. Suppose that you send a particle