All AP Physics 2 Resources
Example Questions
Example Question #8 : Magnetic Fields
An infinitely long wire carries a current of determine the magnitude of the magnetic field away.
None of these
Magnetic field of an infinitely long wire:
Where
Plugging in values:
Example Question #11 : Magnetism And Electromagnetism
A circuit contains a battery and a resistor in series. Determine the magnitude of the magnetic field outside of the loop away from the wire.
None of these
Using
Converting to and plugging in values
Determining current:
Example Question #1 : Magnetic Fields
How strong would a magnetic field need to be in order to make a particle with a mass of and a charge of move in a circular path with a speed of and a radius of ?
For this question, we are being asked to determine the magnetic field necessary to make a particle of a given mass and charge to move in a circular path with a given speed and radius.
To begin with, we can realize that the particle will be moving in a circular path. Thus, there is going to be a centripetal force associated with this circular motion. Moreover, because we know the particle will be present in a magnetic field, we can infer that the magnetic force will be the source of the centripetal force. Thus, we can start by writing out the expression for each of these forces, and then setting them equal to one another.
Rearranging the above expression to isolate the term for magnetic field, we arrive at the following expression.
Now, we can plug in the values given to us in the question stem to solve for the magnetic field strength.
Example Question #11 : Magnetic Fields
Radius of moon:
A loop of current carrying wire runs along the equator of the moon. Determine the magnetic field at the center of the loop if are traveling through it
None of these
Using
Plugging in values
Example Question #11 : Magnetism And Electromagnetism
A particle with a charge of is moving at perpendicularly through a magnetic field with a strength of . What is the magnitude of the force on the particle?
There is no force on the particle
The equation for finding the force on a moving charged particle in a magnetic field is as follows:
Here, is the force in Newtons, is the charge in Coulombs, is the velocity in , and is the magnetic field strength in Teslas.
Another way to write the equation without the cross-product is as follows:
Here, is the angle between the particles velocity and the magnetic field.
For our problem, because theta is , evaluates to 1, so we just need to perform multiplication.
Therefore, the force on the particle is 0.3N.
Example Question #12 : Magnetism And Electromagnetism
An wire with a current of is oriented from parallel to a magnetic field with a strength of . What is the force on the wire?
The wire doesn't experience any force
The equation for the force on a current carrying wire in a magnetic field is as follows:
is the force in Newtons, is the current in amperes, is the magnetic field strength in Teslas, and is the angle from parallel to the magnetic field.
Because our wire is not fully perpendicular to the magnetic field, it does not experience the full possible force. Instead, it experiences times the maximum force value.
Example Question #1 : Magnetic Force
A proton traveling at in a horizontal plane passes through an opening into a mass spectrometer with a uniform magnetic field directed upward. The particle then moves in a circular path through and crashes into the wall of the spectrometer adjacent to the entrance opening. How far down from the entrance is the proton when it crashes into the wall?
The proton’s mass is and its electric charge is .
A charged particle moving through a perpendicular magnetic field feels a Lorentz force equal to the formula:
is the charge, is the particle speed, and is the magnetic field strength. This force is always directed perpendicular to the particle’s direction of travel at that moment, and thus acts as a centripetal force. This force is also given by the equation:
We can set these two equations equal to one another, allowing us to solve for the radius of the arc.
Once the particle travels through a semicircle, it is laterally one diameter in distance from where is started (i.e. twice the radius of the circle).
Twice this value is the lateral offset of its crash point from the entrance:
Example Question #3 : Magnetic Force
What is the force experienced by a charge moving at through a magnetic field with strength at from perpendicular to the field?
There is no net force on the charge
To find the force experienced by the charge, we use this equation:
Because the charge is moving at an angle from perpendicular, we need to take the cross product into account.
Theta is the angle from perpendicular, which is . Plug in known values and solve.
Example Question #4 : Magnetic Force
An charge is moving through a magnetic field at a speed of perpendicular to the direction of the field. What is the force on the charge?
There is no force on the charge
The equation for force on a charge moving through a magnetic field is:
Because the velocity is perpendicular to the field, the cross product doesn't matter, and we can do simple multiplication.
Therefore, the force on the charge is
Example Question #1 : Magnetic Force
A charge is moving through a magnetic field at a speed of from parallel to the magnetic field. What is the force on the charge?
The charge experiences no force
The equation for force on a charge moving through a magnetic field is:
.
The cross product is:
Above, is the degree from parallel the charge is moving. The charge is moving at from parallel, so the equation, once we plug in our numbers, is:
The force on the charge is about .