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By the end of this lesson, students should be able to:
Define a magnetic field.
Calculate the magnetic force on a moving charge.
Determine the direction of magnetic forces using the right-hand rule.
Analyze the motion of charged particles in magnetic fields.
Calculate the radius of circular motion in a magnetic field.
Solve AP Physics C problems involving magnetic forces.
A magnetic field is a region of space in which moving charges experience magnetic forces.
Magnetic fields are represented by:
$$
\vec{B}
$$
The SI unit of magnetic field is the tesla (T).
Magnetic fields can be produced by:
Permanent magnets
Electric currents
Moving charged particles
Unlike electric fields, magnetic fields exert forces only on moving charges.
The magnetic force on a moving charge is given by:
$$
\vec{F}_B=q\vec{v}\times\vec{B}
$$
where:
\(\vec{F}_B\) = magnetic force
\(q\) = charge
\(\vec{v}\) = velocity
\(\vec{B}\) = magnetic field
The symbol \(\times\) indicates a cross product.
The magnitude of the force is:
$$
F_B=|q|vB\sin\theta
$$
where:
\(v\) = speed of the particle
\(B\) = magnetic field strength
\(\theta\) = angle between (\vec{v}) and (\vec{B})
If the velocity is parallel to the magnetic field:
$$
\theta=0^\circ
$$
and
$$
F_B=|q|vB\sin 0^\circ0
$$
No magnetic force acts on the particle.
If:
$$
\theta=180^\circ
$$
then
$$
F_B=|q|vB\sin180^\circ0
$$
Again, no magnetic force exists.
If:
$$
\theta=90^\circ
$$
then
$$
F_B=|q|vB
$$
This produces the maximum possible magnetic force.
The magnetic force is always perpendicular to:
The velocity vector
The magnetic field vector
Therefore:
$$
\vec{F}_B \perp \vec{v}
$$
and
$$
\vec{F}_B \perp \vec{B}
$$
Because the force is perpendicular to the velocity:
$$
W=0
$$
The magnetic force does no work on a charged particle.
Since no work is done:
$$
\Delta K=0
$$
The kinetic energy remains constant.
Therefore:
Speed remains constant.
Direction may change.
This is a critical AP Physics concept.
For positive charges:
Point your fingers in the direction of velocity.
Curl them toward the magnetic field.
Your thumb points in the direction of the magnetic force.
This follows:
$$
\vec{F}_B=q\vec{v}\times\vec{B}
$$
For negative charges:
Use the right-hand rule for a positive charge, then reverse the direction.
Negative charges experience force opposite the right-hand-rule direction.
A magnetic field coming out of the page is represented by:
$$
\odot
$$
The symbol resembles the tip of an arrow coming toward you.
A magnetic field going into the page is represented by:
$$
\otimes
$$
The symbol resembles the tail feathers of an arrow moving away.
Suppose a charged particle enters a uniform magnetic field with:
$$
\vec{v}\perp\vec{B}
$$
The magnetic force is always perpendicular to the velocity.
This force acts as a centripetal force.
The particle follows a circular path.
The magnetic force provides:
$$
F_c=\frac{mv^2}{r}
$$
Set magnetic force equal to centripetal force:
$$
qvB=\frac{mv^2}{r}
$$
Solving for radius:
$$
r=\frac{mv}{|q|B}
$$
The radius increases when:
Mass increases
Speed increases
The radius decreases when:
Charge magnitude increases
Magnetic field strength increases
The circumference is:
$$
2\pi r
$$
The period is:
$$
T=\frac{2\pi r}{v}
$$
Substituting the radius:
$$
T=\frac{2\pi m}{|q|B}
$$
Notice:
$$
T=\frac{2\pi m}{|q|B}
$$
does not depend on speed.
This is a favorite AP exam concept question.
If a particle enters at an angle:
The velocity can be separated into:
$$
v_{\parallel}
$$
and
$$
v_{\perp}
$$
The parallel component remains unchanged.
The perpendicular component produces circular motion.
Combining these motions creates a helix.
A proton moves at:
$$
v=3.0\times10^6m/s
$$
through a magnetic field:
$$
B=0.50T
$$
The velocity is perpendicular to the field.
Find the magnetic force.
Use:
$$
F_B=qvB
$$
Substitute:
$$
F_B=(1.60\times10^{-19})(3.0\times10^6)(0.50)
$$
$$
F_B=2.4\times10^{-13}N
$$
$$
F_B=2.4\times10^{-13}N
$$
An electron moves with speed:
$$
v=2.0\times10^6m/s
$$
in a magnetic field:
$$
B=0.20T
$$
Find the radius of the circular path.
Electron mass:
$$
m=9.11\times10^{-31}kg
$$
Use:
$$
r=\frac{mv}{|q|B}
$$
Substitute:
$$
r=\frac{(9.11\times10^{-31})(2.0\times10^6)}{(1.60\times10^{-19})(0.20)}
$$
$$
r=5.7\times10^{-5}m
$$
$$
r=5.7\times10^{-5}m
$$
Using:
$$
F=qE
$$
instead of:
$$
F=qvB
$$
Electric and magnetic forces are different.
Forgetting the angle factor.
The complete equation is:
$$
F=qvB\sin\theta
$$
Using the right-hand rule incorrectly for negative charges.
Always reverse the direction for negative charges.
Before calculating:
Draw (\vec{v})
Draw (\vec{B})
Determine (\vec{F})
A correct diagram often prevents sign mistakes.
Ask:
Parallel?
Perpendicular?
At an angle?
This immediately determines whether the particle travels in a straight line, circle, or helix.
Since magnetic forces do no work:
$$
K=\text{constant}
$$
Speed remains unchanged.
Only direction changes.
The magnetic force on a moving charge is:
$$
\vec{F}_B=q\vec{v}\times\vec{B}
$$
The magnitude is:
$$
F_B=|q|vB\sin\theta
$$
Magnetic force is always perpendicular to velocity.
Magnetic forces do no work.
$$
W=0
$$
Speed remains constant.
A charge moving perpendicular to a magnetic field undergoes circular motion.
$$
r=\frac{mv}{|q|B}
$$
The period of motion is:
$$
T=\frac{2\pi m}{|q|B}
$$
Charges entering at an angle follow helical paths.
The right-hand rule determines magnetic force direction.
Magnetic forces are fundamental to particle accelerators, mass spectrometers, and many modern technologies.
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