Magnetism and Electromagnetism - AP Physics 2
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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.
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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.
A charged particle, Q, is traveling along a magnetic field, B, with speed v. What is the magnitude of the force the particle experiences?
A charged particle, Q, is traveling along a magnetic field, B, with speed v. What is the magnitude of the force the particle experiences?
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Charged particles only experience a magnetic force when some component of their velocity is perpendicular to the magnetic field. Here, the velocity is parallel to the magnetic field so the particle does not experience a force.
Charged particles only experience a magnetic force when some component of their velocity is perpendicular to the magnetic field. Here, the velocity is parallel to the magnetic field so the particle does not experience a force.
An electron is traveling into a screen at the same time another electron is traveling out of the screen. They are far enough away to not collide with each other. Will the magnetic forces between them be attractive or repulsive?
An electron is traveling into a screen at the same time another electron is traveling out of the screen. They are far enough away to not collide with each other. Will the magnetic forces between them be attractive or repulsive?
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Per conventional symbols, the "X" represents electrons traveling into the page, and the "dot" represents electrons coming out of the page. Picking either electron and using the right hand rule for current traveling, then reversing for the negative charge of the electron gives us the direction of the magnetic field at the other electron. Using the right hand rule again for moving charges in a magnetic field, then reversing for the negative charge of the electron shows that the charges will experience an attractive magnetic force.
Per conventional symbols, the "X" represents electrons traveling into the page, and the "dot" represents electrons coming out of the page. Picking either electron and using the right hand rule for current traveling, then reversing for the negative charge of the electron gives us the direction of the magnetic field at the other electron. Using the right hand rule again for moving charges in a magnetic field, then reversing for the negative charge of the electron shows that the charges will experience an attractive magnetic force.
A conductive rod is moving through a region of magnetic field, directed out of the page as diagrammed above. As a result of its motion, the mobile charge in the rod separates, creating an electric potential across the length of the rod. The length of the rod is 0.12m and the magnitude of the magnetic field is 0.022T. If the rod is moving with velocity 
, what is the magnitude and direction of the potential from one end of the rod to the other?
A conductive rod is moving through a region of magnetic field, directed out of the page as diagrammed above. As a result of its motion, the mobile charge in the rod separates, creating an electric potential across the length of the rod. The length of the rod is 0.12m and the magnitude of the magnetic field is 0.022T. If the rod is moving with velocity
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For a conductor moving in a magnetic field, cutting straight across the field lines, a potential is generated 
where 
is the potential or EMF, 
is the magnetic field strength, and 
is the conductor's velocity relative to the field.
As the rod moves, the positive charge feels an upward-directed force by the right-hand rule, and the negative a downward force resulting in the top being at a higher potential than the bottom of the rod.
For a conductor moving in a magnetic field, cutting straight across the field lines, a potential is generated
As the rod moves, the positive charge feels an upward-directed force by the right-hand rule, and the negative a downward force resulting in the top being at a higher potential than the bottom of the rod.
There is a particle with a charge of 
moving 
perpendicular through a magnetic field with a strength of 
. What is the force on the particle?
There is a particle with a charge of
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The equation for force on a moving charged particle in a magnetic field is
.
Because the charge is moving perpendicularly through the magnetic field, we don't have to worry about the cross product, and the equation becomes simple multiplication.
Therefore, the force experienced by the particle is 3619N.
The equation for force on a moving charged particle in a magnetic field is
Because the charge is moving perpendicularly through the magnetic field, we don't have to worry about the cross product, and the equation becomes simple multiplication.
Therefore, the force experienced by the particle is 3619N.
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
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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.
A conductive rod is moving through a region of magnetic field, as diagrammed above. As a result of its motion, mobile charge carriers in the conductor separate, creating an electric potential across the rod. When, if ever, do the charge carriers cease this motion?
A conductive rod is moving through a region of magnetic field, as diagrammed above. As a result of its motion, mobile charge carriers in the conductor separate, creating an electric potential across the rod. When, if ever, do the charge carriers cease this motion?
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The separated charge creates a potential 
. This potential results in an electric field 
When this induced electric field creates a force 
equal to the magnetic force 
on the mobile charge carriers, motion stops. Of course, if an electric circuit is created drawing current from the rod, motion will resume to rebuild the field.
The separated charge creates a potential
A magnetic rod is inside a coil of wire which is connected to an ammeter. If the rod is stationary, which of the following statements are true?
A magnetic rod is inside a coil of wire which is connected to an ammeter. If the rod is stationary, which of the following statements are true?
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Current is induced in wire when the magnetic flux changes. When the magnetic rod is in motion, the flux is changing, so current is induced. If the coil were expanded or contracted with the rod still there, the flux would change and current would be induced. In our case, the rod is stationary and the coil isn't changing shape. Therefore, the flux is not changing, so there is no current being induced. Additionally, there's no reason for the rod to lose its magnetic property.
Current is induced in wire when the magnetic flux changes. When the magnetic rod is in motion, the flux is changing, so current is induced. If the coil were expanded or contracted with the rod still there, the flux would change and current would be induced. In our case, the rod is stationary and the coil isn't changing shape. Therefore, the flux is not changing, so there is no current being induced. Additionally, there's no reason for the rod to lose its magnetic property.
Suppose the induced electromotive force of a double loop wire has a magnitude of 
when the change in magnetic flux is 
. How much time has elapsed for the flux change?
Suppose the induced electromotive force of a double loop wire has a magnitude of
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Write the formula for induced electromotive force.
Since there are two loops, 
.
Solve for 
.
Write the formula for induced electromotive force.
Since there are two loops,
Solve for
A system consists of two rails, whose resistance is zero, a conductive bar, whose resistance is also zero, and a resistor whose resistance, 0.02 Ohms. The bar remains in contact with the rails as it slides to the right with a velocity 
. The distance from one rail to the other is 0.07m. The magnetic field, which is directed out of the page, has a magnitude of 0.0015T. What is the magnitude and direction of current in the resistor?
A system consists of two rails, whose resistance is zero, a conductive bar, whose resistance is also zero, and a resistor whose resistance, 0.02 Ohms. The bar remains in contact with the rails as it slides to the right with a velocity
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When a conductor moves through a magnetic field in such a way that it cuts through magnetic field lines, the mobile charge carriers separate due to the magnetic force on them, creating a potential 
. Since there is no resistance anywhere else in the circuit, all of this potential is lost in the resistor, so we can apply Ohm's law:
Because the positive charge in the rod feels an upward force due to the right-hand rule, the top of the rod has a greater potential than the bottom, and current flows counterclockwise around the circuit, resulting in a downward direction in the resistor.
When a conductor moves through a magnetic field in such a way that it cuts through magnetic field lines, the mobile charge carriers separate due to the magnetic force on them, creating a potential
Because the positive charge in the rod feels an upward force due to the right-hand rule, the top of the rod has a greater potential than the bottom, and current flows counterclockwise around the circuit, resulting in a downward direction in the resistor.
A 
resistor is connected to a coil with 
turns and a cross sectional area of 
. A magnet is lowered as shown in the figure. The magnetic field increases from 
to 
in 
. Find the magnitude of the current going through the resistor.
A
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There is a change in flux produced by the changing magnetic field which is given by
where 
is the cross sectional area of the coil, 
is the change in magnetic field, and 
is the angle of the field lines relative to the normal of the cross section of the coil.
In this case the magnet is perpendicular to the cross section of the coil and so
The change in magnetic field is just the final given value minus the intial value. Faraday's Law says that an emf will be generated by a change in flux,
where 
is the number of turns in the coil. Plugging in the change in flux gives
The change in time is just 
since we can start our clock at zero. The current can be found using Ohm's Law where
There is a change in flux produced by the changing magnetic field which is given by
where
In this case the magnet is perpendicular to the cross section of the coil and so
The change in magnetic field is just the final given value minus the intial value. Faraday's Law says that an emf will be generated by a change in flux,
where
The change in time is just
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
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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.
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.
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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:
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
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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:
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
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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.
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
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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.
An infinitely long wire carries a current of 
determine the magnitude of the magnetic field 
away.
An infinitely long wire carries a current of
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Magnetic field of an infinitely long wire:
Where 
Plugging in values:
Magnetic field of an infinitely long wire:
Where
Plugging in values:
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
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Using
Plugging in values
Using
Plugging in values
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
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Using
Converting 
to 
and plugging in values
Determining current:
Using
Converting
Determining current:
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
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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.