Question

A horizontally oriented coil of wire of radius \(5.00 \mathrm{~cm}\) and carrying a current, \(i\), is being levitated by the south pole of a vertically oriented bar magnet suspended above the center of the coil. If the magnetic field on all parts of the coil makes an angle \(\theta\) of \(45.0^{\circ}\) with the vertical, determine the magnitude and the direction of the current needed to keep the coil floating in midair. The magnitude of the magnetic field is \(B=0.0100 \mathrm{~T}\), the number of turns in the coil is \(N=10.0\), and the total coil mass is \(10.0 \mathrm{~g}\).

Short answer

Answer: The magnitude of the current needed is \(6.34 \mathrm{~A}\), and the current should flow in a clockwise direction when viewed from above.

Step-by-step solution

Calculate the Gravitational Force

To calculate the force of gravity acting on the coil, we can use the formula: $$F_g = m \cdot g$$ where \(F_g\) is the gravitational force, \(m\) is the mass of the coil, and \(g\) is the acceleration due to gravity (approximately \(9.81 \mathrm{~m/s^2}\)). Given the mass \(m = 10.0 \mathrm{~g} = 0.010 \mathrm{~kg}\), we can calculate the gravitational force: $$F_g = 0.010 \mathrm{~kg} \times 9.81 \mathrm{~m/s^2} = 9.81 \times 10^{-2} \mathrm{N}$$

Calculate the Magnetic Force

The magnetic force acting on the coil can be calculated using the formula: $$F_m = B \cdot I \cdot L \cdot \sin(\theta)$$ where \(F_m\) is the magnetic force, \(B\) is the magnetic field, \(I\) is the current, \(L\) is the length of the wire, and \(\theta\) is the angle between the magnetic field and the vertical direction. In our case, \(B = 0.0100 \mathrm{~T}\), \(\theta = 45.0^{\circ}\), and since there are \(N = 10.0\) turns in the coil, the total length of the wire in the field is $$L = 2\pi r \cdot N = 2\pi \cdot 0.05 \mathrm{~m} \cdot 10.0 = \pi \mathrm{~m}$$

Equate Magnetic Force to Gravitational Force

To keep the coil levitated, we need the magnetic force to be equal to the gravitational force: $$F_m = F_g$$ Using the magnetic force formula, we have: $$B \cdot I \cdot L \cdot \sin(\theta) = 9.81 \times 10^{-2} \mathrm{N}$$ Now, we can solve for the current \(I\): $$I = \frac{9.81 \times 10^{-2} \mathrm{N}}{0.0100 \mathrm{~T} \cdot \pi \mathrm{~m} \cdot \sin(45.0^{\circ})}$$

Calculate the Current

Now, let's plug in the values and find the magnitude of the current needed: $$I = \frac{9.81 \times 10^{-2} \mathrm{N}}{0.0100 \mathrm{~T} \cdot \pi \mathrm{~m} \cdot \frac{\sqrt{2}}{2}} = \frac{9.81 \times 10^{-2} \mathrm{N} \cdot 2 \sqrt{2}}{0.0100 \mathrm{~T} \cdot \pi \mathrm{~m}}$$ $$I = 6.34 \mathrm{~A}$$

Determine the Direction of the Current

To determine the direction of the current, we need to consider the orientation of the magnetic field and the levitating force. Since the south pole of the magnet is above the coil, the magnetic field lines are directed downwards. By using the Right Hand Rule, we can determine that the current needs to flow in a clockwise direction when viewed from above for the magnetic force to act upwards and to counteract the gravitational force. So, to keep the coil floating in midair, the required current is \(6.34 \mathrm{~A}\) flowing in a clockwise direction.

Key concepts

Magnetic Force

Magnetic force is the force exerted by a magnetic field on a charged particle or current-carrying conductor. In the context of magnetic levitation, this force is a crucial component to counteract gravitational force and achieve levitation.

Magnetic force (F_m ) is calculated using the formula: \( F_m = B \cdot I \cdot L \cdot \sin(\theta) \), where:

• \( B \) represents the magnetic field strength (in teslas).
• \( I \) is the current (in amperes).
• \( L \) is the length of the conductor in the magnetic field.
• \( \theta \) is the angle between the magnetic field and the current's direction.

This formula highlights that the magnetic force depends on both the strength and direction of the magnetic field, as well as the amount of current and the effective length of the conductor.

In our example, the coil must generate enough magnetic force to levitate by matching the gravitational pull, thus demanding precise calculations of these contributing factors.

Right Hand Rule

The Right Hand Rule is a simple method used to determine the direction of the magnetic force on a current-carrying conductor. It is vital in situations like magnetic levitation, where both direction and magnitude of forces need to be carefully managed.

To apply the Right Hand Rule, remember:

• Point your thumb in the direction of the current (I).
• Extend your fingers in the direction of the magnetic field (B).
• Your palm will face the direction of the resulting force (F_m).

This convenient tool helps to visualize and predict how a current loop will interact with external magnetic fields.

In the scenario presented, since the magnetic field lines are directed downward from the magnet's south pole, the Right Hand Rule informs us that the current must flow clockwise to produce an upward magnetic force.

Magnetic Field

A magnetic field is an invisible field that exerts magnetic forces on moving charges and magnetized objects. In this case, it plays a central role in magnetic levitation. The strength of a magnetic field is measured in teslas (T). It determines how much force the field can exert on charges within its influence.

Several components characterize a magnetic field:

• Strength (B): The intensity of the magnetic influence, here given as 0.0100 T.
• Direction: Orientation relative to the object, herein at an angle of 45 degrees with the vertical.

Each element contributes to the overall environment in which the coil operates.

This specific setup aims to achieve a balance, requiring the coil to generate an opposing force necessary for levitation without physically touching the magnet.

Gravitational Force

Gravitational force is the attractive force exerted by the Earth on any object with mass. It acts downward and is pivotal to the principles of levitation because it is what must be counteracted to achieve a floating effect.

The gravitational force (\( F_g \)) in the example problem is calculated as \( F_g = m \cdot g \), where:

• \( m \) is the mass of the object in kilograms.
• \( g \) is the acceleration due to gravity (approximately 9.81 \(m/s^2\)).

For the given coil with mass 10.0 grams (0.010 kg), this results in a gravitational force of approximately 0.0981 N.

Understanding the gravitational pull on the coil is essential, as it sets the benchmark for the magnetic force necessary to achieve levitation. The goal is to generate a magnetic force equal in magnitude but opposite in direction to this gravitational force.