Question

A copper wire of radius \(0.500 \mathrm{~mm}\) is carrying a current at the Earth's Equator. Assuming that the magnetic field of the Earth has magnitude 0.500 G at the Equator and is parallel to the surface of the Earth and that the current in the wire flows toward the east, what current is required to allow the wire to levitate?

Short answer

Answer: The current required for the copper wire to levitate at the Earth's Equator is 1.38 A.

Step-by-step solution

Find the cross-sectional area of the wire

To calculate the force due to gravity, we need to find the mass of the wire, which requires us to know its cross-sectional area A. The wire has a circular cross section, so we use the formula: \(A = \pi r^2\), where r is the radius. A = π (0.0005 m)^2 = 7.854 × 10^-7 m^2

Find the mass of the wire per unit length

To find the mass per unit length of the wire, we'll need to use the volume of the wire (cross-sectional area x length), and then multiply by the density of copper (\(\rho_{cu} = 8.96 \times 10^3 kg/m^3\)). Mass per unit length (m) = ρ (A x L) / L = ρA m = (8.96 x 10^3 kg/m^3)(7.854 x 10^-7 m^2) = 7.041 x 10^-3 kg/m

Write down the gravitational force and magnetic force equations

The force of gravity (Fg) is given by the equation Fg = mg, where g is the acceleration due to gravity (approx. 9.81 m/s^2). Fg = (7.041*10^-3 kg/m)(9.81 m/s^2) = 6.901*10^-2 N/m The magnetic force (Fm) is given by the equation Fm = iBw, where i is the current in the wire, B is the magnetic field, and w is the width of the wire (perpendicular to the direction of the current).

Set the gravitational force equation equal to the magnetic force equation

For the wire to levitate, the magnetic force must be equal to the gravitational force: Fm = Fg iBw = mg

Solve for the current i in the wire

Now we solve for the current i: i = mg / (Bw) As given, Earth's magnetic field, \(B = 0.500 G = 5.0 \times 10^{-5} T\) (converting Gauss to Tesla). Since the current flows towards the East, the magnetic force will be applied in an upwards direction; therefore, the width of the wire, w, can be assumed equal to the diameter, which is \(2 \times 0.0005 m = 0.001 m\) . i = (6.901 × 10^-2 N/m) / ((5.0 × 10^-5 T)(0.001 m)) = 1.38 A. Therefore, a current of 1.38 A is required for the copper wire to levitate at the Earth's Equator, due to the Earth's magnetic field.

Key concepts

Magnetic Force

The concept of magnetic force is central to understanding how magnetic levitation works. This force arises when a current-carrying conductor, like a copper wire, is placed within a magnetic field. Here, the Earth’s magnetic field interacts with the electrical current to create a lifting force.

Let's break this down: when a current flows through a wire, it generates its own magnetic field around the wire. Now, if this field interacts with another magnetic field—for instance, Earth’s magnetic field—a force is exerted on the wire. This force is what we term as the magnetic force. It follows the left-hand rule, meaning if the thumb of your left hand is pointing in the direction of the current, and your fingers point in the direction of the magnetic field, your palm would face the direction of the force.

In order for the wire to levitate, this magnetic force must be strong enough to counteract the wire's weight. The formula to calculate the magnetic force, as given in the solution, is Fm = iBw, where i is the current, B is the magnetic field, and w is the width of the wire.

Gravitational Force

While the magnetic force is pulling the wire upwards, the gravitational force is simultaneously pulling it downwards towards the earth. This force is due to the Earth’s gravity, which acts on all objects with mass.

The gravitational force is determined by the equation Fg = mg, where m is the mass of the object—or in this case, the mass per unit length of the wire—and g is the acceleration due to gravity (approximately 9.81 m/s² on Earth).

In magnetic levitation, the goal is for the magnetic force to precisely balance this gravitational pull. When these two forces are equal, the net force on the wire is zero, allowing it to levitate. As shown in the step-by-step solution, we calculate the mass per unit length of the wire and then use the Fg formula to determine the gravitational force acting on it.

Current and Magnetic Fields

The relationship between current and magnetic fields is described by Ampère’s Law, which states that for any closed loop path, the sum of the length elements times the magnetic field in the direction of the length element is equal to the permeability times the electric current enclosed in the loop.

It's essential to understand that the magnetic field produced by a current-carrying wire isn’t isolated; it interacts with other magnetic fields. In our exercise, the wire's magnetic field interacts with the Earth's magnetic field. The direction of the current (towards the east), in conjunction with the direction of Earth's magnetic field (parallel to the surface), determines the direction of the magnetic force—upwards in this case, as required for levitation.

Moreover, the size of the current will determine the magnitude of the magnetic field around the wire, and consequently, the magnitude of the magnetic force. An increase in current results in a stronger magnetic force and vice versa.

Density of Copper

Lastly, the density of copper plays a significant role in determining whether the copper wire can levitate. Density is mass per unit volume, which for copper is about 8.96 x 10³ kg/m³. The wire’s density is intrinsic to finding both its mass and gravitational force.

Since the wire’s ability to levitate depends on the magnetic force equaling the gravitational pull, knowing the density of copper helps us calculate the wire's mass for any given length. By multiplying the wire's cross-sectional area by its density, we obtain the mass per unit length (m).

Understanding the density is crucial because it directly affects the mass and therefore the gravitational force that must be balanced by the magnetic force for levitation to occur. In the problem, by knowing the density of copper, we can deduce the mass of the wire per meter, and using the aforementioned formulas, we can solve for the current needed for levitation.