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(a) What is the speed of a supersonic aircraft with a 17.0-m wingspan, if it experiences a 1.60-V Hall voltage between its wing tips when in level flight over the north magnetic pole, where the Earth's field strength is \(8.00 \times 10^{-5} \mathrm{~T} ?\) (b) Explain why very little current flows as a result of this Hall voltage.

Short Answer

Expert verified
The speed of the supersonic aircraft is approximately 1180.2 m/s. Very little current flows due to the high electrical resistance of air.

Step by step solution

01

Identify the Relevant Concepts

The Hall effect is the production of a voltage difference (the Hall voltage) across an electrical conductor, transverse to an electric current in the conductor and a magnetic field perpendicular to the current. The Hall voltage is given by the formula: \( V_H = B \times l \times v \), where \( V_H \) is the Hall voltage, \( B \) is the magnetic field strength, \( l \) is the distance between the voltage points (the wingspan in this case), and \( v \) is the speed of the aircraft. We’ll use this formula to solve part (a).
02

Calculate the Speed of the Aircraft (Part a)

Rearrange the formula to solve for speed (v): \( v = \frac{V_H}{B \times l} \). Substitute the given values into the formula: \( v = \frac{1.60}{8.00 \times 10^{-5} \times 17.0} \) meters per second.
03

Perform the Calculation

Carry out the calculation from step 2: \( v = \frac{1.60}{8.00 \times 10^{-5} \times 17.0} \approx 1180.2 \) meters per second.
04

Address the Current Flow (Part b)

Very little current flows as a result of the Hall voltage because the electrical conductivity of air is very low, and hence its resistance is very high. So, even if a voltage is induced between the wing tips, the circuit is essentially open and cannot sustain a constant current flow.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hall Voltage
Hall voltage is a fascinating phenomenon that plays a critical role in various electronic devices. It is a voltage difference produced across a conductor when it is subject to a magnetic field perpendicular to the direction of the electric current flowing through it. The Hall voltage arises due to the Hall effect, named after Edwin Hall, who discovered it in 1879.

Imagine electrons moving through a conductor—when a magnetic field is applied transversely, these charged particles experience a force called the Lorentz force, which pushes them to one side of the conductor. As a result, a potential difference is created, and this is what we measure as the Hall voltage. The magnitude of this voltage is directly proportional to the strength of the magnetic field, the amount of electric current, and inversely proportional to the charge carrier density in the conductor.

Understanding Hall Voltage in Practice

In the exercise's example, a supersonic aircraft with a wingspan acting as the 'conductor' experiences a voltage (the Hall voltage) between its wingtips. Using the Hall effect formula, \( V_H = B \times l \times v \), where \( B \) is the magnetic field strength, \( l \) is the distance between the wingtips, and \( v \) is the velocity of the aircraft, we can deduce the speed of the craft by rearranging and solving for \( v \).
Magnetic Field Strength
The magnetic field strength, often denoted by \( B \), quantifies how strong a magnetic field is. Its unit of measure is the Tesla (T). Magnetic fields are intrinsic in the Earth and in magnets, and they can also be generated artificially through electric currents and magnetic materials. The field's strength is a key determinant in the Hall effect, influencing the magnitude of the Hall voltage generated.

Calculating Speed Using Magnetic Field Strength

For instance, the magnetic field strength at the north magnetic pole (\(8.00 \times 10^{-5} \mathrm{~T}\)), is essential information required to solve part (a) of the exercise. It is used in conjunction with the Hall voltage and wingspan to determine the velocity of the aircraft. Magnetic field strength tells us about the environment in which the aircraft is operating and is an indispensable variable in applying the Hall effect formula for practical calculations.
Electrical Conductivity
Electrical conductivity is an indication of a material's ability to conduct electric current. When a voltage is applied to a material, it will respond by allowing a certain amount of current to pass through it—this property is what we define as electrical conductivity. It plays a crucial role in determining the behavior of materials when exposed to electric and magnetic fields.

Materials with high electrical conductivity, like metals, allow electrons to move through them freely, facilitating the flow of current. Conversely, materials with low electrical conductivity, such as air or rubber, resist the free flow of electrons, meaning that they are insulators.

Role of Electrical Conductivity in the Hall Effect

In part (b) of our exercise, we find that little current flows as a result of the Hall voltage because the air has a low electrical conductivity. Even though the aircraft's wingspan establishes a potential difference, due to the high resistance and low conductivity of the surrounding air, the circuit is effectively open, and current cannot flow continuously. Understanding a material's electrical conductivity is crucial when explaining phenomena such as the Hall effect in real-life applications.

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Most popular questions from this chapter

(a) A 0.750-m-long section of cable carrying current to a car starter motor makes an angle of \(60^{\circ}\) with the Earth's \(5.50 \times 10^{-5} \mathrm{~T}\) field. What is the current when the wire experiences a force of \(7.00 \times 10^{-3} \mathrm{~N} ?(\mathrm{~b})\) If you run the wire between the poles of a strong horseshoe magnet, subjecting \(5.00 \mathrm{~cm}\) of it to a \(1.75-T\) field, what force is exerted on this segment of wire?

(a) What is the force per meter on a lightning bolt at the equator that carries 20,000 A perpendicular to the Earth's \(3.00 \times 10^{-5}-\mathrm{T}\) field? (b) What is the direction of the force if the current is straight up and the Earth's field direction is due north, parallel to the ground?

(a) Triply charged uranium-235 and uranium-238 ions are being separated in a mass spectrometer. (The much rarer uranium-235 is used as reactor fuel.) The masses of the ions are \(3.90 \times 10^{-25} \mathrm{~kg}\) and \(3.95 \times 10^{-25} \mathrm{~kg}\), respectively, and they travel at \(3.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\) in a 0.250-T field. What is the separation between their paths when they hit a target after traversing a semicircle? (b) Discuss whether this distance between their paths seems to be big enough to be practical in the separation of uranium-235 from uranium-238.

Nonnuclear submarines use batteries for power when submerged. (a) Find the magnetic field \(50.0 \mathrm{~cm}\) from a straight wire carrying \(1200 \mathrm{~A}\) from the batteries to the drive mechanism of a submarine. (b) What is the field if the wires to and from the drive mechanism are side by side? (c) Discuss the effects this could have for a compass on the submarine that is not shielded.

(a) A DC power line for a light-rail system carries \(1000 \mathrm{~A}\) at an angle of \(30.0^{\circ}\) to the Earth's \(5.00 \times 10^{-5}-\mathrm{T}\) field. What is the force on a \(100-\mathrm{m}\) section of this line? (b) Discuss practical concerns this presents, if any.

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