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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

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