Magnetism and Electromagnetism - AP Physics 2

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.

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.

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.

Write the formula for induced electromotive force.

Since there are two loops, .

Solve for .

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.

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

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 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.

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 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:

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.

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.

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.

Magnetic field of an infinitely long wire:

Where

Plugging in values:

Using

Plugging in values

Using

Converting to and plugging in values

Determining current:

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.

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:

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.