Determining the force on a charge moving in a B-field

Since a moving charge produces a magnetic field it should come as no surprise that a moving charge placed in an external magnetic field will feel a magnetic force. (Because of the pole law).
∙Furthermore, a stationary charge in a magnetic field will feel no magnetic force because the charge will not have its own magnetic field.
∙In fact, the force F felt by a charge q traveling at velocity v through a B-field of strength B is given by

∙The direction of F is given by another right hand rule.
∙Place the heel of your right hand in the plane containing v and B so that your curled fingertips touch v first:
∙Your extended thumb points in the direction of the force on a (+) charge.

FYI
∙F is perpendicular to v and B and is thus perpendicular to the plane of v and B.
∙F is in the opposite direction for a (-) charge.

PRACTICE:

Consider a charge \( q \) traveling at velocity \( v \) in the magnetic field \( B \) shown here.
Show that \( r = \frac{mv}{qB} \).

SOLUTION:

 Since \( v \) is in the blue plane and \( B \) points out toward you, \( v \perp B \) and \( \sin 90^\circ = 1 \).
 Thus \( F = qvB \).
 But \( F = ma \), so that \( qvB = ma \).
 Since the charge is in UCM, then \( a = \frac{v^2}{r} \).
 Thus \( qvB = \frac{mv^2}{r} \implies r = \frac{mv}{qB} \).

EXAMPLE:

The tendency of a moving charge to follow a curved trajectory in a magnetic field is used in a mass spectrometer.

∙An unknown element is ionized, and accelerated by an applied voltage in the chamber S.
∙It strikes a phosphorescent screen and flashes.
∙By measuring x, one can determine the mass of the ion, and hence the unknown element.

Force on a current-carrying conductor in a B-field

∙We already know the effect of a magnetic field on a moving charge.
∙It stands to reason that a wire carrying a current in a magnetic field will also feel a force, because current is moving charge.
∙A wire with no current feels no magnetic force:
∙But a wire with a current will be deflected by a magnetic force as shown:
∙Knowing the RHR for a charge is all you need to determine the direction of the force in the wire.

Force on a current-carrying conductor in a B-field

∙We now know the direction of the magnetic force acting on a current-carrying wire if it is in a magnetic field.
∙The magnitude of the magnetic force F acting on a wire of length L and carrying a current of I in a magnetic field B is given by this formula:

FYI
Note that the direction of I is also the direction of q as it flow through the wire

EXAMPLE:

Beginning with the formula \( F = qvB \sin \theta \), show that \( F = BIL \sin \theta \).

SOLUTION:
1. \( F = qvB \sin \theta \) \(\quad\) (given)
2. \( F = q \left(\frac{L}{t}\right) B \sin \theta \) \(\quad\) (\( v = \frac{\text{distance}}{\text{time}} \))
3. \( F = \left(\frac{q}{t}\right) LB \sin \theta \) \(\quad\) (just move the \( t \))
4. \( F = ILB \sin \theta \) \(\quad\) (\( I = \frac{\text{charge}}{\text{time}} \))
5. \( F = BIL \sin \theta \) \(\quad\) (commutative property)

EXAMPLE:

James Clerk Maxwell developed the theory that showed that the electric field and the magnetic field were manifestations of a single force called the electromagnetic force. Both the electromagnetic force and the gravitational force travel as waves through space at the speed of light. Compare and contrast the two waves.

SOLUTION:

∙The effect of a gravitational disturbance on an object is to stretch and shrink it in time with the wave.

EXAMPLE:

Find the value of \( \frac{1}{\sqrt{\mu_0 \epsilon_0}} \).

SOLUTION:
 The permittivity of free space is \( \epsilon_0 = 8.85 \times 10^{-12} \).
 The permeability of free space is \( \mu_0 = 4\pi \times 10^{-7} \).
 Then:

\[
\frac{1}{\sqrt{\mu_0 \epsilon_0}} = \frac{1}{\sqrt{(8.85 \times 10^{-12})(4\pi \times 10^{-7})}}
= 2.998 \times 10^8 \, \text{ms}^{-1}.
\]

FYI:

Thus \( c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \).

EXAMPLE:

How are the electric field and the magnetic field related in electromagnetic radiation (light)?

SOLUTION:

Observe the animation:
∙They are perpendicular, and they are in phase.