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 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 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 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.

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.

Write the formula for inductance and substitute the givens. The length of the solenoid has no effect in this problem.

The relationship between the number of turns for the primary coil and secondary coil in a transformer ( and respectively) to the relative potentials is

Solving for ,

We need to use the right hand rule to solve this problem. However, the right hand rule applies to the flow of current, which is in the opposite direction of the actual flow of electrons (current is defined, in this case, as the direction of proton flow). Therefore, you can either use the right hand rule and reverse what you determine, or simply use your left hand.

Let's just use our left hand. Point your thumb out and curl your fingers. Your fingers should be pointing at you. This is the direction of the electric field when electrons are traveling the direction of your thumb. If you lay your left thumb along the wire loop on the left side of the loop, our fingers are inside the loop and pointing out of the screen. This is the scenario we are looking for. Therefore, the electrons need to flow clockwise around the loop.

Note that the electrons must flow through the wire, eliminating the answer options for "into the screen" and "out of the screen."

Let's use the right-hand rule to determine the magnetic field cause by each current.

For current , we determine that the magnetic field is going into the screen.

For current , we determine that the magnetic field is going out of the screen.

Do the two directions cancel out? Well, the magnitude of is greater than the magnitude of , meaning that will overpower , so the net direction is out of the screen.

Using the right hand rule, we can tell that the direction of the magnetic field due to the bottom wire is out of the screen. Likewise, we can tell that the magnetic field due to the top wire is into the screen. Because point R is halfway between the two wires, they each have the same strength. Therefore, they both cancel each other out, leaving no magnetic field.

Recall that the convention for the direction of current is from the positive end of the voltage source to the negative end (opposite the direction of flow of electrons). Thus, in this circuit the current is flowing counterclockwise from the voltage source. Using the right hand rule for the conventional current in the wire, the right thumb is pointed along the wire pointing to the left. At point the fingers curl around and point up, out of the screen. This can be verified by putting the thumb in the direction of current anywhere in the circuit. For example, if we take the direction of the current across the resistor, our thumb points down. Curling our fingers around the wire, the fingers will again point out of the screen at point , verifying our initial answer.

Current flows counterclockwise in this circuit. Using the right hand rule for the conventional current in the wire, the right thumb is pointed along the wire pointing to the left at the top of the circuit. At point the fingers curl around and point down, into the screen.

The current flows through this circuit counterclockwise. Using the right hand rule for the conventional current in the wire, the right thumb is pointed along the wire pointing to the right in the wire at the bottom of the circuit. At point the fingers curl around and point down, into the screen.

Visualizing a globe and pointing the thump "south" towards the "north" magnetic pole shows that the fingers curl and point from east to west.

Write the formula for inductance and substitute the givens. The length of the solenoid has no effect in this problem.

The relationship between the number of turns for the primary coil and secondary coil in a transformer ( and respectively) to the relative potentials is

Solving for ,

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.

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.