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.

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."

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 :

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 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.

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.

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 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 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: