
By the end of this lesson, students should be able to:
Explain why a current-carrying wire experiences a magnetic force.
Calculate the magnetic force on a straight wire in a magnetic field.
Determine the direction of the magnetic force using the right-hand rule.
Analyze the motion of current-carrying conductors in magnetic fields.
Understand the operation of electric motors.
Solve AP Physics C problems involving magnetic forces on wires.
In the previous lesson, we learned that a moving charge experiences a magnetic force:
$$
\vec{F}_B=q\vec{v}\times\vec{B}
$$
A current in a wire consists of many moving charges.
Therefore, a current-carrying wire placed in a magnetic field also experiences a magnetic force.
This idea forms the basis of:
Electric motors
Loudspeakers
Galvanometers
Electromagnetic actuators
Consider a straight wire carrying current (I) in a uniform magnetic field (\vec{B}).
The magnetic force on the wire is:
$$
\vec{F}=I\vec{L}\times\vec{B}
$$
where:
\(\vec{F}\) = magnetic force on the wire
\(I\) = current
\(\vec{L}\) = length vector of the wire
\(\vec{B}\) = magnetic field
The direction of (\vec{L}) is the direction of conventional current.
The magnitude is:
$$
F=ILB\sin\theta
$$
where:
\(I\) = current
\(L\) = length of wire in the field
\(B\) = magnetic field strength
\(\theta\) = angle between the current direction and the magnetic field
If:
$$
\theta=0^\circ
$$
then:
$$
F=ILB\sin0^\circ=0
$$
No magnetic force acts on the wire.
If:
$$
\theta=180^\circ
$$
then:
$$
F=ILB\sin180^\circ=0
$$
Again, the force is zero.
If:
$$
\theta=90^\circ
$$
then:
$$
F=ILB
$$
The magnetic force is maximum.
Inside the wire, charge carriers move with drift velocity.
Each moving charge experiences a magnetic force:
$$
\vec{F}_B=q\vec{v}\times\vec{B}
$$
The sum of all these microscopic forces produces a net force on the wire itself.
The magnetic field pushes on the moving charges.
The charges collide with atoms in the wire.
As a result, the entire wire experiences a force.
To determine the direction of force:
Point your fingers in the direction of current.
Curl them toward the magnetic field.
Your thumb points in the direction of the force.
This follows:
$$
\vec{F}=I\vec{L}\times\vec{B}
$$
Always use:
Conventional current direction
Not electron flow direction
when applying the right-hand rule.
Represented by:
$$
\odot
$$
The dot represents the tip of an arrow coming toward you.
Represented by:
$$
\otimes
$$
The cross represents the tail of an arrow moving away from you.
Only the portion of wire actually inside the magnetic field experiences force.
If only a section of wire is in the field:
Use that length in:
$$
F=ILB\sin\theta
$$
If a 2.0 m wire exists but only 0.50 m lies within the field:
Use:
$$
L=0.50m
$$
in calculations.
Consider a rectangular current loop in a magnetic field.
Two sides experience forces in opposite directions.
These forces create a torque.
The resulting torque causes the loop to rotate.
This principle forms the basis of electric motors.
The torque is:
$$
\tau=N I A B \sin\theta
$$
where:
\(N\) = number of turns
\(I\) = current
\(A\) = loop area
\(B\) = magnetic field
\(\theta\) = angle between the area vector and field
An electric motor converts:
Electrical Energy
into
Mechanical Energy
using magnetic forces on current-carrying conductors.
A simple motor contains:
Current loop
Magnetic field
Power source
Commutator
The magnetic forces generate continuous rotation.
A wire carries:
$$
I=5.0A
$$
through a magnetic field:
$$
B=0.40T
$$
The wire length in the field is:
$$
L=0.60m
$$
The wire is perpendicular to the field.
Find the magnetic force.
Use:
$$
F=ILB
$$
Substitute:
$$
F=(5.0)(0.40)(0.60)
$$
$$
F=1.2N
$$
$$
F=1.2N
$$
A wire carries:
$$
I=10A
$$
through a magnetic field:
$$
B=0.25T
$$
The length is:
$$
L=0.80m
$$
The angle between the current and field is:
$$
30^\circ
$$
Find the magnetic force.
Use:
$$
F=ILB\sin\theta
$$
Substitute:
$$
F=(10)(0.80)(0.25)\sin30^\circ
$$
Since:
$$
\sin30^\circ=0.5
$$
$$
F=1.0N
$$
$$
F=1.0N
$$
A wire experiences a force:
$$
F=3.0N
$$
while in a magnetic field:
$$
B=0.50T
$$
The wire length is:
$$
L=1.5m
$$
The wire is perpendicular to the field.
Find the current.
Use:
$$
F=ILB
$$
Solve for current:
$$
I=\frac{F}{LB}
$$
Substitute:
$$
I=\frac{3.0}{(1.5)(0.50)}
$$
$$
I=4.0A
$$
$$
I=4.0A
$$
Forgetting the angle factor.
The complete equation is:
$$
F=ILB\sin\theta
$$
not simply:
$$
F=ILB
$$
unless the wire is perpendicular to the field.
Using electron flow direction.
Always use conventional current when applying the right-hand rule.
Using the total wire length.
Only the section inside the magnetic field contributes to the force.
Before calculating:
Draw the current.
Draw the magnetic field.
Determine the force direction.
This avoids most sign errors.
Always ask:
Parallel?
Perpendicular?
General angle?
This often simplifies the calculation immediately.
When current loops appear:
Think about:
Opposing forces
Torque
Rotation
Many AP conceptual questions focus on motor operation.
A current-carrying wire experiences a magnetic force because its charges are moving.
The magnetic force on a wire is:
$$
\vec{F}=I\vec{L}\times\vec{B}
$$
The magnitude is:
$$
F=ILB\sin\theta
$$
Maximum force occurs when:
$$
\theta=90^\circ
$$
No force occurs when:
$$
\theta=0^\circ
$$
or
$$
\theta=180^\circ
$$
Force direction is determined using the right-hand rule.
Current loops experience torque:
$$
\tau=N I A B \sin\theta
$$
Magnetic forces on wires form the operating principle behind electric motors and many electromagnetic devices.
Understanding magnetic forces on wires is essential before studying magnetic fields produced by currents and electromagnetic induction.
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