Magnetism - AP Physics C: Electricity and Magnetism
Card 1 of 154
An infinitely long wire has a current of 
running through it. Calculate the magnetic field at a distance 
away from the wire.
An infinitely long wire has a current of
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For infinitely long wires, the formula for the magnetic field is 
, where 
is the current and 
is the distance from the wire.
The magnetic field is calculated using our given values.
For infinitely long wires, the formula for the magnetic field is
The magnetic field is calculated using our given values.
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A solenoid is 
long and it is made up of 
turns of wire. How much current must run through the solenoid to generate a magnetic field of 
inside of the solenoid?
A solenoid is
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The formula for the magnetic field inside the solenoid is 
, where 
is the number of turns of wire, 
is the length of the solenoid, and 
is the current.
We want to find the current so we solve for 
.
Plug in the values.
The formula for the magnetic field inside the solenoid is
We want to find the current so we solve for
Plug in the values.
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Two parallel wires a distance 
apart each carry a current 
, and repel each other with a force 
per unit length. If the current in each wire is doubled to 
, and the distance between them is halved to 
, by what factor does the force per unit length change?
Two parallel wires a distance
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Relevant equations:
Step 1: Find the original and new magnetic fields created by wire 1 at wire 2, using Ampere's law with an Amperian loop of radius 
or 
, respectively.
Original
New
Since the wires are parallel to each other and wire 1's field is directed circularly around it, in each case wire 1's field is perpendicular to wire 2.
Step 2: Find the original and new magnetic forces per unit length on wire 2, due to the field created by wire 1.
Original
New
So, the new force per unit length is 8 times greater than the original.
Relevant equations:
Step 1: Find the original and new magnetic fields created by wire 1 at wire 2, using Ampere's law with an Amperian loop of radius
Original
New
Since the wires are parallel to each other and wire 1's field is directed circularly around it, in each case wire 1's field is perpendicular to wire 2.
Step 2: Find the original and new magnetic forces per unit length on wire 2, due to the field created by wire 1.
Original
New
So, the new force per unit length is 8 times greater than the original.
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A region of uniform magnetic field, 
, is represented by the grey area of the box in the diagram. The magnetic field is oriented into the page.
A stream of protons moving at velocity 
is directed into the region of the magnetic field, as shown. Identify the correct path of the stream of protons after they enter the region of magnetic field.
A region of uniform magnetic field,
A stream of protons moving at velocity
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The magnetic force on a moving charged particle is given by the equation:
Isolating the directional component of this equation yields the understanding that the resulting force on a moving charged particle is perpendicular to the plane of the velocity vector and magnetic field vector. Using the right-hand-rule on this cross-product shows that the velocity vector right-crossed into the magnetic field vector into the page yields a magnetic force vector upward on a positive charge. This will result in a semi-circular path oriented vertically upward.
The magnetic force on a moving charged particle is given by the equation:
Isolating the directional component of this equation yields the understanding that the resulting force on a moving charged particle is perpendicular to the plane of the velocity vector and magnetic field vector. Using the right-hand-rule on this cross-product shows that the velocity vector right-crossed into the magnetic field vector into the page yields a magnetic force vector upward on a positive charge. This will result in a semi-circular path oriented vertically upward.
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Two infinitely long wires having currents 
and 
are separated by a distance 
.
The current 
is 6A into the page. The current 
is 9A into the page. The distance of separation is 1.5mm. The point 
lies 1.5mm away from 
on a line connecting the centers of the two wires.
What is the magnitude and direction of the net magnetic field at the point 
?
Two infinitely long wires having currents
The current
What is the magnitude and direction of the net magnetic field at the point
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At point 
, the magnetic field due to 
points right (via the right hand rule) with a magnitude given by:
At point 
, the magnetic field due to 
points right (via the right hand rule) with a magnitude given by:
The addition of these two vectors, both pointing in the same direction, results in a net magnetic field vector of magnitude 
to the right.
At point
At point
The addition of these two vectors, both pointing in the same direction, results in a net magnetic field vector of magnitude
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Consider a current-carrying loop with current 
, radius 
, and center 
.
What is the direction of the magnetic field produced?
Consider a current-carrying loop with current
What is the direction of the magnetic field produced?
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The correct answer is into the page. As the current is moving clockwise, we can use our right hand rule for magnetic fields produced by a current-carrying loop. Curl the fingers of your right hand in the direction of the current. This should result in your thumb pointing toward the screen, indicating the direction of the magnetic field.
The correct answer is into the page. As the current is moving clockwise, we can use our right hand rule for magnetic fields produced by a current-carrying loop. Curl the fingers of your right hand in the direction of the current. This should result in your thumb pointing toward the screen, indicating the direction of the magnetic field.
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Consider a current-carrying loop with current 
, radius 
, and center 
.
What is the magnitude of the magnetic field at point 
?
Consider a current-carrying loop with current
What is the magnitude of the magnetic field at point
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The current flowing clockwise through the wire will induce a magentic field directed into the screen. The magnitude of such a magnetic field is given by the equation:
The current flowing clockwise through the wire will induce a magentic field directed into the screen. The magnitude of such a magnetic field is given by the equation:
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Consider a current-carrying loop with current 
, radius 
, and center 
.
What would happen to the magnetic field at point 
if the radius was halved and current was multiplied by four?
Consider a current-carrying loop with current
What would happen to the magnetic field at point
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The current flowing clockwise through the wire will induce a magentic field directed into the screen. The magnitude of such a magnetic field is given by the equation:
Using out altered values, we can derive a ratio to determine the change in magnetic field.
The resulting field will be eight times stronger than the original.
The current flowing clockwise through the wire will induce a magentic field directed into the screen. The magnitude of such a magnetic field is given by the equation:
Using out altered values, we can derive a ratio to determine the change in magnetic field.
The resulting field will be eight times stronger than the original.
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Consider two long, straight, current-carrying wires at distance 
from each other, each with a current of magnitude 
going in opposite directions.
What is the magnitude of the magnetic field at a point equidistant from both wires?
Consider two long, straight, current-carrying wires at distance
What is the magnitude of the magnetic field at a point equidistant from both wires?
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Using our right hand rule for magnetic fields produced by current-carrying wires, we know that the magnetic field produced by each wire is in the same direction within the distance between the wires. Therefore, we know that the total magnetic field is simply the magnetic field of one of the wires multiplied by two.
Use the equation for magnetic field:
Multiply by two, since the magnetic field will be equal for each wire, and substitute the given variables:
Using our right hand rule for magnetic fields produced by current-carrying wires, we know that the magnetic field produced by each wire is in the same direction within the distance between the wires. Therefore, we know that the total magnetic field is simply the magnetic field of one of the wires multiplied by two.
Use the equation for magnetic field:
Multiply by two, since the magnetic field will be equal for each wire, and substitute the given variables:
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A proton traveling 
enters a uniform magnetic field and experiences a magnetic force, causing it to travel in a circular path. Taking the magnetic field to be 
, what is the radius of this circular path (shown in red)?
A proton traveling
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To calculate the magnetic force of a single charge, we use 
, where 
is the charge of the proton, 
is its velocity, 
is the uniform magnetic field.
Since this magnetic force causes the proton to travel in a circular path, we set this magnetic force equation equal to the centripetal force equation.
is the mass of the proton and 
is the radius of the circular path. Solve for 
.
Using the values given in the question, we can solve for the radius.
To calculate the magnetic force of a single charge, we use
Since this magnetic force causes the proton to travel in a circular path, we set this magnetic force equation equal to the centripetal force equation.
Using the values given in the question, we can solve for the radius.
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Which of the following best describes the net magnetic flux through a closed sphere, in the presence of a magnet?
Which of the following best describes the net magnetic flux through a closed sphere, in the presence of a magnet?
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The net magnetic flux (or net field flowing in and out) through any closed surface must always be zero. This is because magnetic field lines have no starting or ending points, so any field line going into the surface must also come out. In other words, "there are no magnetic monopoles."
The net magnetic flux (or net field flowing in and out) through any closed surface must always be zero. This is because magnetic field lines have no starting or ending points, so any field line going into the surface must also come out. In other words, "there are no magnetic monopoles."
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A particle of charge 
and mass 
moves with a speed of 
perpendicular to a uniform magnetic field, 
. What is the period of the particle's orbit in the field?
A particle of charge
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Relevant equations:
Set the magnetic force equal to the centripetal force, since the magnetic force is directed towards the center of the particle's circular path and centripetal force is defined as the net force towards the center of a circular path.
Rearrange to isolate the velocity:
Determine the distance, 
, traveled in one revolution, which is the circumference of a circle of radius 
:
Plug this distance and velocity into 
, to solve for the period 
:
Relevant equations:
Set the magnetic force equal to the centripetal force, since the magnetic force is directed towards the center of the particle's circular path and centripetal force is defined as the net force towards the center of a circular path.
Rearrange to isolate the velocity:
Determine the distance,
Plug this distance and velocity into
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Consider a current-carrying loop with current 
, radius 
, and center 
.
A particle with charge 
flies through the center and into the page with velocity 
. What is the total electromagnetic force on the particle at the instant that it flies through the loop, in terms of the variables given?
Consider a current-carrying loop with current
A particle with charge
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The correct answer is zero. To calculate the force of a magnetic field on a moving charged particle, we use the cross product. We know that if the magnetic field is parallel to the velocity vector of the particle, then the force produced is zero.
Because our magnetic field in this case is going in the same direction as the velocity of the particle, we know that the magnetic force on the particle is zero.
The correct answer is zero. To calculate the force of a magnetic field on a moving charged particle, we use the cross product. We know that if the magnetic field is parallel to the velocity vector of the particle, then the force produced is zero.
Because our magnetic field in this case is going in the same direction as the velocity of the particle, we know that the magnetic force on the particle is zero.
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Consider two long, straight, current-carrying wires at distance 
from each other, each with a current of magnitude 
going in opposite directions.
If the two wires described are not held in place, what motion will result from the magnetic fields produced?
Consider two long, straight, current-carrying wires at distance
If the two wires described are not held in place, what motion will result from the magnetic fields produced?
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The answer is that the wires will move from each other. Using our right hand rule, we know that the magnetic fields produced by each wire are in the same direction, as long as their currents oppose. Using the right hand rule again to determine the direction of the force exerted on each wire by the magnetic field with which they are interacting yields a force in the direction away from the other wire for each wire.
The answer is that the wires will move from each other. Using our right hand rule, we know that the magnetic fields produced by each wire are in the same direction, as long as their currents oppose. Using the right hand rule again to determine the direction of the force exerted on each wire by the magnetic field with which they are interacting yields a force in the direction away from the other wire for each wire.
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A proton traveling enters a uniform magnetic field and experiences a magnetic force, causing it to travel in a circular path. Taking the magnetic field to be , what is the radius of this circular path (shown in red)?
A proton traveling
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To calculate the magnetic force of a single charge, we use , where is the charge of the proton, is its velocity, is the uniform magnetic field.
Since this magnetic force causes the proton to travel in a circular path, we set this magnetic force equation equal to the centripetal force equation.
is the mass of the proton and is the radius of the circular path. Solve for .
Using the values given in the question, we can solve for the radius.
To calculate the magnetic force of a single charge, we use
Since this magnetic force causes the proton to travel in a circular path, we set this magnetic force equation equal to the centripetal force equation.
Using the values given in the question, we can solve for the radius.
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Which of the following best describes the net magnetic flux through a closed sphere, in the presence of a magnet?
Which of the following best describes the net magnetic flux through a closed sphere, in the presence of a magnet?
Tap to reveal answer
The net magnetic flux (or net field flowing in and out) through any closed surface must always be zero. This is because magnetic field lines have no starting or ending points, so any field line going into the surface must also come out. In other words, "there are no magnetic monopoles."
The net magnetic flux (or net field flowing in and out) through any closed surface must always be zero. This is because magnetic field lines have no starting or ending points, so any field line going into the surface must also come out. In other words, "there are no magnetic monopoles."
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A particle of charge and mass moves with a speed of perpendicular to a uniform magnetic field, . What is the period of the particle's orbit in the field?
A particle of charge
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Relevant equations:
Set the magnetic force equal to the centripetal force, since the magnetic force is directed towards the center of the particle's circular path and centripetal force is defined as the net force towards the center of a circular path.
Rearrange to isolate the velocity:
Determine the distance, , traveled in one revolution, which is the circumference of a circle of radius :
Plug this distance and velocity into , to solve for the period :
Relevant equations:
Set the magnetic force equal to the centripetal force, since the magnetic force is directed towards the center of the particle's circular path and centripetal force is defined as the net force towards the center of a circular path.
Rearrange to isolate the velocity:
Determine the distance,
Plug this distance and velocity into
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Consider a current-carrying loop with current , radius , and center .
A particle with charge flies through the center and into the page with velocity . What is the total electromagnetic force on the particle at the instant that it flies through the loop, in terms of the variables given?
Consider a current-carrying loop with current
A particle with charge
Tap to reveal answer
The correct answer is zero. To calculate the force of a magnetic field on a moving charged particle, we use the cross product. We know that if the magnetic field is parallel to the velocity vector of the particle, then the force produced is zero.
Because our magnetic field in this case is going in the same direction as the velocity of the particle, we know that the magnetic force on the particle is zero.
The correct answer is zero. To calculate the force of a magnetic field on a moving charged particle, we use the cross product. We know that if the magnetic field is parallel to the velocity vector of the particle, then the force produced is zero.
Because our magnetic field in this case is going in the same direction as the velocity of the particle, we know that the magnetic force on the particle is zero.
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Consider two long, straight, current-carrying wires at distance from each other, each with a current of magnitude going in opposite directions.
If the two wires described are not held in place, what motion will result from the magnetic fields produced?
Consider two long, straight, current-carrying wires at distance
If the two wires described are not held in place, what motion will result from the magnetic fields produced?
Tap to reveal answer
The answer is that the wires will move from each other. Using our right hand rule, we know that the magnetic fields produced by each wire are in the same direction, as long as their currents oppose. Using the right hand rule again to determine the direction of the force exerted on each wire by the magnetic field with which they are interacting yields a force in the direction away from the other wire for each wire.
The answer is that the wires will move from each other. Using our right hand rule, we know that the magnetic fields produced by each wire are in the same direction, as long as their currents oppose. Using the right hand rule again to determine the direction of the force exerted on each wire by the magnetic field with which they are interacting yields a force in the direction away from the other wire for each wire.
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A proton traveling enters a uniform magnetic field and experiences a magnetic force, causing it to travel in a circular path. Taking the magnetic field to be , what is the radius of this circular path (shown in red)?
A proton traveling
Tap to reveal answer
To calculate the magnetic force of a single charge, we use , where is the charge of the proton, is its velocity, is the uniform magnetic field.
Since this magnetic force causes the proton to travel in a circular path, we set this magnetic force equation equal to the centripetal force equation.
is the mass of the proton and is the radius of the circular path. Solve for .
Using the values given in the question, we can solve for the radius.
To calculate the magnetic force of a single charge, we use
Since this magnetic force causes the proton to travel in a circular path, we set this magnetic force equation equal to the centripetal force equation.
Using the values given in the question, we can solve for the radius.
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