AP Physics 2 Exam Question: Magnetic Fields

A circular loop with radius R and resistance R is connected in series with a battery with emf ε as shown below:

The loop is placed in a uniform magnetic field B directed out of the plane of the loop. The loop is initially at rest. Answer the following questions:

(a) Derive an expression for the magnetic field strength B in terms of the applied voltage V, the resistance R, and the dimensions of the loop.

(b) Assuming that at time t = 0, the battery is connected such that a current I starts flowing clockwise through the loop, determine the magnitude and direction of the magnetic force acting on the loop at time t > 0.

(c) After some time, the loop reaches a terminal velocity v due to the presence of a retarding magnetic force. Determine an expression for the terminal velocity v in terms of the applied voltage V, the resistance R, and the dimensions of the loop.

(d) If the resistance R of the loop is doubled while keeping all other parameters constant, how does the terminal velocity change? Explain your answer.

Solution:

(a) To derive an expression for the magnetic field strength B, we use Ohm's Law and the relationship between voltage, resistance, and current. The induced emf ε in the loop is given by:

ε = B * (2πR) * v,

where v is the velocity of the loop and (2πR) is the circumference of the loop. Since the resistance R is the same as the applied voltage V (Ohm's Law), we can rewrite the equation as:

V = B * (2πR) * v.

Rearranging the equation, we can solve for B:

B = V / (2πRv).

(b) The magnitude of the magnetic force F on a current-carrying loop in a magnetic field is given by:

|F| = |BIL|,

where I is the current flowing through the loop and L is the length of the loop. The force is perpendicular to both the magnetic field and the current.

In our case, the current I is given by:

I = V / R (using Ohm's Law).

Substituting the given values, the magnitude of the magnetic force F is:

|F| = |(V / (2πRv)) * (V / R) * (2πR)|.

Simplifying the expression, we get:

|F| = |V^2 / v|.

The direction of the force can be determined using the right-hand rule. Since the current flows clockwise, the force will be directed out of the page (opposite to the magnetic field).

(c) The retarding magnetic force is equal to the electrostatic force created by the induced emf. At terminal velocity, the net force on the loop is zero. Therefore, we have:

Retarding magnetic force = Electrostatic force.

Using the expression for the electrostatic force:

|Fe| = |(V / R)|,

And the expression for the retarding magnetic force:

|Fm| = |V^2 / v|,

Setting them equal to each other, we get:

|V / R| = |V^2 / v|.

Simplifying the expression, we find:

v = V^2 / R.

(d) If the resistance R of the loop is doubled while keeping all other parameters constant, the terminal velocity v would decrease. This is because the retarding magnetic force is inversely proportional to the resistance. When the resistance is doubled, the magnitude of the retarding magnetic force is halved, leading to a decrease in the terminal velocity.

Note: It's important to note that this explanation assumes a simplified scenario where the loop is ideal and there are no other factors like gravitational force or changes in the magnetic field strength.