
By the end of this lesson, students should be able to:
Explain how electric currents create magnetic fields.
Determine the direction of magnetic fields around current-carrying wires.
Apply the Right-Hand Rule for magnetic fields.
Calculate magnetic fields produced by long straight wires.
Analyze magnetic fields produced by circular loops and solenoids.
Solve AP Physics C problems involving magnetic fields generated by currents.
Previously, we learned that magnetic fields exert forces on moving charges and current-carrying wires.
An equally important fact is that electric currents themselves produce magnetic fields.
This discovery was made in 1820 by Hans Christian Ørsted when he observed that a compass needle deflected near a current-carrying wire.
This observation revealed a fundamental connection between electricity and magnetism.
Consider a long straight wire carrying current (I).
The magnetic field produced by the wire forms concentric circles around the wire.
The field lines:
Circle the wire.
Become weaker farther from the wire.
Have no beginning or end.
Unlike electric field lines, magnetic field lines always form closed loops.
To determine the magnetic field direction:
Point your thumb in the direction of conventional current.
Curl your fingers around the wire.
Your fingers indicate the direction of the magnetic field.
This rule is extremely important in AP Physics C.
Represent the current by:
$$
\odot
$$
The magnetic field circles counterclockwise.
Represent the current by:
$$
\otimes
$$
The magnetic field circles clockwise.
The magnetic field produced by a long straight wire is:
$$
B=\frac{\mu_0 I}{2\pi r}
$$
where:
\(B\) = magnetic field
\(I\) = current
\(r\) = distance from the wire
\(\mu_0\) = permeability of free space
The constant:
$$
\mu_0=4\pi\times10^{-7};T\cdot m/A
$$
is a fundamental constant in electromagnetism.
Notice:
$$
B\propto\frac{1}{r}
$$
As distance increases:
Magnetic field decreases.
The decrease is slower than the electric field of a point charge.
Compare:
Electric field:
$$
E\propto\frac{1}{r^2}
$$
Magnetic field of a wire:
$$
B\propto\frac{1}{r}
$$
When multiple currents are present:
The total magnetic field equals the vector sum of individual magnetic fields.
Mathematically:
$$
\vec{B}_{total}=\sum \vec{B}
$$
This principle is essential for solving multi-wire problems.
A current-carrying wire creates a magnetic field.
A nearby wire carrying current experiences a magnetic force due to that field.
Thus:
Wire 1 produces a field.
Wire 2 experiences a force.
The interaction is mutual.
If two parallel wires carry currents in the same direction:
They attract.
If two parallel wires carry currents in opposite directions:
They repel.
For two long parallel wires:
$$
\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi d}
$$
where:
\(F\) = force
\(L\) = wire length
\(I_1\) = first current
\(I_2\) = second current
\(d\) = separation distance
This interaction was once used to define the ampere.
A circular current loop produces a magnetic field at its center.
The magnitude is:
$$
B=\frac{\mu_0 I}{2R}
$$
where:
\(R\) = radius of the loop
For (N) turns:
$$
B=\frac{\mu_0 N I}{2R}
$$
The field increases proportionally with the number of turns.
Curl your fingers in the direction of current.
Your thumb points in the direction of the magnetic field through the center of the loop.
A solenoid is a long coil of wire carrying current.
Examples include:
Electromagnets
Relays
MRI machines
Inductors
For an ideal solenoid:
$$
B=\mu_0 n I
$$
where:
\(n\) = number of turns per meter
\(I\) = current
Inside an ideal solenoid:
Field is nearly uniform.
Field lines are parallel.
Magnetic field resembles the electric field between parallel plates.
$$
B=\frac{\mu_0 I}{2\pi r}
$$
Field lines form circles.
$$
B=\frac{\mu_0 I}{2R}
$$
Field points through the center.
$$
B=\mu_0 n I
$$
Field is approximately uniform inside.
A long wire carries:
$$
I=8.0A
$$
Find the magnetic field at:
$$
r=0.040m
$$
Use:
$$
B=\frac{\mu_0 I}{2\pi r}
$$
Substitute:
$$
B=\frac{(4\pi\times10^{-7})(8.0)}{2\pi(0.040)}
$$
Simplify:
$$
B=4.0\times10^{-5}T
$$
$$
B=4.0\times10^{-5}T
$$
A circular loop has:
$$
I=5.0A
$$
and
$$
R=0.10m
$$
Find the magnetic field at the center.
Use:
$$
B=\frac{\mu_0 I}{2R}
$$
Substitute:
$$
B=\frac{(4\pi\times10^{-7})(5.0)}{2(0.10)}
$$
$$
B=3.14\times10^{-5}T
$$
$$
B=3.14\times10^{-5}T
$$
A solenoid has:
$$
n=1500
;turns/m
$$
and
$$
I=2.0A
$$
Find the magnetic field inside.
Use:
$$
B=\mu_0 n I
$$
Substitute:
$$
B=(4\pi\times10^{-7})(1500)(2.0)
$$
$$
B=3.77\times10^{-3}T
$$
$$
B=3.77\times10^{-3}T
$$
Confusing the right-hand rule for force with the right-hand rule for field direction.
Remember:
Current → Field direction
Velocity → Force direction
These are different applications.
Using the wrong distance dependence.
For a straight wire:
$$
B\propto\frac{1}{r}
$$
not
$$
\frac{1}{r^2}
$$
Forgetting superposition.
When multiple wires exist:
Add magnetic fields as vectors.
Before calculating:
Sketch the current.
Use the right-hand rule.
Determine field directions.
This often reveals whether fields add or cancel.
Straight wire:
$$
B=\frac{\mu_0 I}{2\pi r}
$$
Loop center:
$$
B=\frac{\mu_0 I}{2R}
$$
Solenoid:
$$
B=\mu_0 n I
$$
These are among the most frequently tested magnetic field equations in AP Physics C.
Electric currents create magnetic fields.
Magnetic field direction is found using the right-hand rule.
Long straight wires produce circular magnetic field lines.
$$
B=\frac{\mu_0 I}{2\pi r}
$$
Circular loops produce fields through their centers.
$$
B=\frac{\mu_0 I}{2R}
$$
Solenoids create nearly uniform magnetic fields.
$$
B=\mu_0 n I
$$
Parallel currents:
Same direction → attract
Opposite directions → repel
Magnetic fields obey superposition.
$$
\vec{B}_{total}=\sum\vec{B}
$$
Current-generated magnetic fields form the foundation for electromagnets, motors, generators, transformers, and many modern technologies.
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