Relevant equations:

Write the expression for the flux using the given expression for the magnetic field strength:

Take the derivative of this flux to find the induced emf:

Determine the maximum value of , and replace  with this value, to find the maximum induced emf:

Plug this maximum induced emf into , to find the maximum induced current:

The magnetic field due to the long straight wire points into the page (right-hand rule) in the region below it in the picture. The magnitude of this magnetic field is given by:

The magnetic field thus decreases in value with increasing distance, . As such, the amount of flux passing through the wire loop decreases steadily as it is moved away from the wire.

According to Lenz's Law, an induced current within the wire loop will counteract this loss of magnetic flux from the straight wire. The magnetic field from the induced current in the wire loop will consequently point in the same direction as that from the current in the straight wire. The right-hand rule for a magnetic field into the page yields an induced current direction in the clockwise direction.

Since there are no dissipative circuit elements (such as a resistor) present in the circuit, the energy originally held by the electric field of the capacitor oscillates between electric energy and magnetic energy over time. As such, the maximum electric energy of the capacitor is equal in value to the maximum magnetic energy of the inductor.

Magnetic flux is a measure of the amount of magnetic field passing through a given area and is given by the equation:

The magnetic field is constant for this problem, and the areas are all circular loops. As such, the vector product integral simplifies to:

The angle theta is the angle between a vector normal to the surface of the area  and the given magnetic field.

For loops C and D, theta is 90 degrees, so the magnetic flux for both of these loops is equal to zero.

For loops A and B, the angle theta is 0 degrees, so the magnetic flux for both of these loops is nonzero and proportional to the area of each loop. Therefore, the flux for loop B is larger than that for loop A because loop B has the larger radius.

The relationship between capacitance, distance, and area is . We can rearrange this equation to solve for area.

Now, we can use the values given in teh question to solve.

The electric potential difference created between the plates of a parallel plate capactor is given by the equation:

The charge can be calculated by using the equation:

The value of the capacitance is related to the dimensions of the capacitor with the equation:

Combining these equations yields:

The area becomes inconsequential, while the potential is directly proportional to the surface charge density and the distance between the plates, and inversely proportional to the dielectric of the material between the plates. Changing the area does not cause any change in the potential difference measured between the plates, and changing any of the other variables would cause a resultant change in the potential difference.

For parallel plate capacitors, the equation is .

Solve for .

Now we can plug in our given values.

Capacitance is related to plate area and distance by the equation .

Given the area and distance, we can solve for capacitance. The voltage, in this case, is irrelevant.

Charge on a capacitor is given by the equation . We know that the voltage is , but we need to determine the capacitance based off of the area and distance between the plates.

We can plug this value into the equation for charge.

The electric field given by a capacitor is given by the formula . We do not have these variables, so we will have to adjust the equation.

The capacitance can be determined by the area of the plates and the distance between them.

This equation simplifies to , allowing us to solve using the values given in the question.