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### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000

# (a) What is the speed of a supersonic aircraft with a ${\mathbf{17}}{\mathbf{.}}{\mathbf{0}}{\mathbf{‐}}{\mathbf{m}}$ wingspan, if it experiences a ${\mathbf{1}}{\mathbf{.}}{\mathbf{60}}{\mathbf{‐}}{\mathbf{V}}$ Hall voltage between its wing tips when in level flight over the north magnetic pole, where the Earth’s field strength is ${\mathbf{8}}{\mathbf{.}}{\mathbf{00}}{\mathbf{×}}{{\mathbf{10}}}^{\mathbf{‐}\mathbf{5}}{\mathbf{}}{\mathbf{T}}$ (b) Explain why very little current flows as a result of this Hall voltage.

(a) The speed of the aircraft is $1.17×{10}^{3}\mathrm{m}/\mathrm{s}$.

(b) The little number of current flows is a result of this Hall voltage as there exists a magnetic force pushing the charges in the opposite direction.

See the step by step solution

## Step 1: Given information

The wingspan of the supersonic aircraft is $\mathrm{L}=17.0\mathrm{m}$

The magnetic field strength of the earth is $\mathrm{B}=8.00×{10}^{‐5}\mathrm{T}$

Induced voltage due to the Hall effect is $\mathrm{E}=1.60\mathrm{V}$

## Step 2: Define the Hall effect

An induced voltage generated by a charge carrier moving in an applied magnetic field perpendicular to the direction of motion can be expressed as,

${\mathrm{E}}{}{=}{\mathrm{BLv}}$ …………………(1)

## Step 2: Evaluating the hall voltage

Substitute the given data in equation (1) to evaluate the speed that produces such hall voltage can be given as,

$\mathrm{v}=\frac{\mathrm{E}}{\mathrm{BL}}\phantom{\rule{0ex}{0ex}}=\frac{1.60\mathrm{V}}{\left(8.00×{10}^{‐5}\mathrm{T}\right)×\left(17.0\mathrm{m}\right)}\phantom{\rule{0ex}{0ex}}=1.17×{10}^{3}\mathrm{m}/\mathrm{s}$

Therefore the speed of supersonic aircraft is $1.17×{10}^{3}\mathrm{m}/\mathrm{s}$.

## Step 3: Explanation of why little current flow as a result of this hall voltage

(b)

Very little current flows as a result of this Hall voltage due to the reason that there exists a magnetic force that pushes the charges in the opposite direction.

Therefore, the flow of a small current is due to the Hall voltage as there already exists a magnetic force pushing the charges in the opposite direction.