Magnetic Fields - AP Physics 2
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Suppose that a proton moves perpendicularly through a magnetic field at a speed of 
. If this proton experiences a magnetic force of 
, what is the strength of the magnetic field?
.
Suppose that a proton moves perpendicularly through a magnetic field at a speed of
To solve this question, we need to relate the speed and charge of the particle with the magnetic force it experiences in order to solve for the magnetic field strength. Thus, we'll need to use the following equation:
Also, we are told that the particle is moving perpendicularly to the magnetic field.
Rearrange to solve for the magnetic field, then plug in known values and solve.
To solve this question, we need to relate the speed and charge of the particle with the magnetic force it experiences in order to solve for the magnetic field strength. Thus, we'll need to use the following equation:
Also, we are told that the particle is moving perpendicularly to the magnetic field.
Rearrange to solve for the magnetic field, then plug in known values and solve.
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Suppose that a positively charged particle with charge 
moves in a circular path of radius 
in a constant magnetic field of strength 
. If the magnetic field strength is doubled to 
, what effect does this have on the radius of the circular path that this charge takes?
Suppose that a positively charged particle with charge
To answer this question, we need to realize that the particle is moving in a circular path because of some sort of centripetal force. Since the charge is moving while within a constant magnetic field, we can conclude that it is the magnetic force that is responsible for the centripetal force that keeps this charge moving in a circle. Thus, we need to relate the centripetal force to the magnetic force.
The above equation shows us that the radius of the circular path is directy proportional to the mass and velocity of the particle, and inversely proportional to the charge of the particle and the magnetic field strength. Thus, if the value of the magnetic field is doubled, the above equation predicts that the value of the radius would be cut in half.
To answer this question, we need to realize that the particle is moving in a circular path because of some sort of centripetal force. Since the charge is moving while within a constant magnetic field, we can conclude that it is the magnetic force that is responsible for the centripetal force that keeps this charge moving in a circle. Thus, we need to relate the centripetal force to the magnetic force.
The above equation shows us that the radius of the circular path is directy proportional to the mass and velocity of the particle, and inversely proportional to the charge of the particle and the magnetic field strength. Thus, if the value of the magnetic field is doubled, the above equation predicts that the value of the radius would be cut in half.
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loops of current carrying wire form a solenoid of length 
that carries 
and have radius 
. Determine the magnetic field at the center of the solenoid.
Using:
Where:
is the magnetic field
is the number of coils
is the current in the solenoid
is the length of the solenoid
is 
Plugging in values:
Using:
Where:
Plugging in values:
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There is a loop with a radius of 
and a current of 
. Determine the magnitude of the magnetic field at the center of the loop.
There is a loop with a radius of
Using the Biot-Savart law:
Where 
is the radius of the loop
is the current
is the distance from the center of the loop
Plugging in values:
Using the Biot-Savart law:
Where
Plugging in values:
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A circular circuit is powered by a 
battery. How will the magnetic field change if the battery is removed and placed in the opposite direction?
A circular circuit is powered by a
Reversing the battery will reverse the direction of the current. Using the right hand rule, it can be seen that this will also reverse the direction of the magnetic field. Since the magnitude of the current stays the same, the magnitude of the magnetic field will as well.
Reversing the battery will reverse the direction of the current. Using the right hand rule, it can be seen that this will also reverse the direction of the magnetic field. Since the magnitude of the current stays the same, the magnitude of the magnetic field will as well.
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A circular circuit is powered by a 
battery. How will the magnetic field change if a second 
battery is added in the same direction as the first?
A circular circuit is powered by a
Based on the Biot-Savart law:
Doubling the voltage will double the current, which will double the magnetic field. The direction will stay the same.
Based on the Biot-Savart law:
Doubling the voltage will double the current, which will double the magnetic field. The direction will stay the same.
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If the north end of a magnetic points towards the geographic north pole, that means that the geographic north pole is a magnetic __________ pole.
If the north end of a magnetic points towards the geographic north pole, that means that the geographic north pole is a magnetic __________ pole.
Magnets will align themselves with the surrounding magnetic field. Thus, if the north pole of a magnet is pointing north, the direction of the magnetic field must be pointing north. Magnetic fields point towards magnetic south poles, so the geographic north pole is actually a magnetic south pole.
Magnets will align themselves with the surrounding magnetic field. Thus, if the north pole of a magnet is pointing north, the direction of the magnetic field must be pointing north. Magnetic fields point towards magnetic south poles, so the geographic north pole is actually a magnetic south pole.
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An infinitely long wire carries a current of 
determine the magnitude of the magnetic field 
away.
An infinitely long wire carries a current of
Magnetic field of an infinitely long wire:
Where 
Plugging in values:
Magnetic field of an infinitely long wire:
Where
Plugging in values:
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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
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
Using
Plugging in values
Using
Plugging in values
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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.
A circuit contains a
Using
Converting 
to 
and plugging in values
Determining current:
Using
Converting
Determining current:
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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 
?
How strong would a magnetic field need to be in order to make a particle with a mass 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.
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.
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