Question

Two identical coaxial coils of wire of radius \(20.0 \mathrm{~cm}\) are directly on top of each other, separated by a 2.00 -mm gap. The lower coil is on a flat table and has a current \(i\) in the clockwise direction; the upper coil carries an identical current and has a mass of \(0.0500 \mathrm{~kg} .\) Determine the magnitude and the direction that the current in the upper coil has to have to keep it levitated at the distance \(2.00 \mathrm{~mm}\) above the lower coil.

Short answer

Answer: The magnitude of the current in the upper coil needed to keep it levitated 2.00 mm above the lower coil is \(\frac{i}{8.28 \cdot 10^3 A}\), and the direction is counterclockwise.

Step-by-step solution

Calculate the magnetic field due to the current in the lower coil

We can determine the magnetic field produced by the lower coil using Ampere's Law. The magnetic field B inside a solenoid carrying a current i is given by: \(B = \mu_0 n i\) Where \(\mu_0\) is the permeability of free space (\(4\pi \cdot 10^{-7} Tm/A\)), \(n\) is the number of turns per unit length, and \(i\) is the current in the coil. We are not given the number of turns per unit length, but since the two coils are identical, we only need to know the relative ratio of the current flowing in both coils in the next steps. Therefore, we can denote the magnetic field produced by the lower coil as \(B_1 = k i\), where \(k = \mu_0 n\) and \(i\) is the current.

Calculate the magnetic force between the two coils

To find the magnetic force between the two coils, we first need to find the magnetic field due to the current in the upper coil. Using the same formula as before and denoting the magnetic field produced by the upper coil as \(B_2 = k i_2\), where \(i_2\) is the current in the upper coil. The force between two coaxial coils carrying currents can be calculated using the formula: \(F_{12} = \frac{μ_0 i_1 i_2 a^2}{2πr^3}\) Where \(i_1\) and \(i_2\) are the currents in the two coils, \(a\) is the radius of the coils, and \(r\) is the distance between the coils.

Balance the gravitational force and the magnetic force

In order to keep the upper coil levitated 2.00 mm above the lower coil, the magnetic force between the coils must be equal to the gravitational force acting on the upper coil. The gravitational force can be calculated using the formula \(F_g = mg\), where \(m\) is the mass of the upper coil and \(g\) is the acceleration due to gravity (approximately \(9.81 m/s^2\)). Setting the magnetic force equal to the gravitational force, we get: \(\frac{μ_0 i i_2 a^2}{2πr^3} = mg\) Rearranging the equation to solve for the current in the upper coil \(i_2\), we get: \(i_2 = \frac{2πr^3 mg}{μ_0 i a^2}\)

Calculate the magnitude and direction of the current in the upper coil

Substitute the given values into the equation to find \(i_2\): \(i_2 = \frac{2π(0.002 m)^3 (0.0500 kg)(9.81 m/s^2)}{(4\pi \cdot 10^{-7} Tm/A) i (0.20 m)^2}= \frac{i}{8.28 \cdot 10^3 A}\) The direction of \(i_2\) must be opposite to the direction of the current in the lower coil to generate a repulsive force, so the current must flow in the counterclockwise direction. Thus, the magnitude of the current in the upper coil needed to keep it levitated 2.00 mm above the lower coil is \(\frac{i}{8.28 \cdot 10^3 A}\), and the direction is counterclockwise.

Key concepts

Ampere's Law

Understanding magnetic fields begins with Ampere's Law, which relates electric currents to the magnetic fields they produce. Ampere's Law tells us that for any closed loop, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability of free space times the total electric current enclosed by the loop. Mathematically, it's expressed as ewline \[ \oint B \cdot dl = \mu_0 I_{enc} \]
where \( B \) is the magnetic field, \( dl \) is a small segment of the loop, and \( I_{enc} \) is the current passing through the enclosed area of the loop.
In our exercise, we used a simplified form of this law, as we sought the magnetic field produced by a single loop of current. Even without the details like the exact number of turns in the coil or their specific arrangement, we understand that an increase in current will proportionally increase the magnetic field due to Ampere's Law. This proportional relationship is crucial in designing systems where magnetic fields are used to exert forces, as is the case with magnetic levitation.

Magnetic Field

The magnetic field is a fundamental aspect of electromagnetism, which intuitively can be thought of as the influence created by magnets or electric currents which can exert a magnetic force on moving charges or other magnets. It's often represented by field lines emanating from the north to the south pole of a magnet, or in a circular pattern around a current-carrying conductor.
In our levitation problem, we measure the magnetic field in teslas (T) and compute its effect based on the geometry of the coils and the current through them. The solenoid's magnetic field is uniform and parallel inside the coil and weaker outside, which is ideal for stable levitation purposes. The uniform field interacts with other magnetic fields to generate forces that can be used for levitation or movement.

Gravitational Force

The gravitational force is a universal force of attraction that acts between all masses. According to Newton's law of universal gravitation, every point mass attracts every other point mass with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers. This force is represented by the equation ewline \[ F_g = m \times g \]
where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity, which is approximately \( 9.81 \text{m/s}^2 \) near the Earth's surface. In the context of our levitation problem, the gravitational force must be balanced by an equal and opposite magnetic force to maintain stable levitation, otherwise the coil would either fall under gravity or be propelled upwards.

Magnetic Force

The magnetic force plays a central role in the phenomenon of magnetic levitation. This force arises from the interaction between a magnetic field and a magnetic object or a second magnetic field. The formula ewline \[ F = \frac{\mu_0 i_1 i_2 a^2}{2\pi r^3} \]
quantifies the force between two current-carrying coils. Here, \( i_1 \) and \( i_2 \) represent the currents in each coil, \( a \) is the radius of the coils, \( r \) is the separation between them, and \( \mu_0 \) is the permeability of free space. Notice how this force is dependent on the magnitude of currents and the distance between the coils.
To achieve levitation as described in the exercise, the magnetic force must be delicately tuned to exactly counteract gravitational force. This is done by adjusting the current in the upper coil until the magnetic repulsion balances the weight of the coil. We also consider that opposite currents will repel each other to determine the correct direction for the current in the levitating coil.