Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

In the August 1992 space shuttle flight, only 250 m of the conducting tether considered in Example 23.2 could be let out. A 40.0 V motional emf was generated in the Earth's $5.00 \times 10^{- 5} T$ field, while moving at $7.80 \times 10^{3} m / s$ . What was the angle between the shuttle’s velocity and the Earth’s field, assuming the conductor was perpendicular to the field?

The angle between the shuttle’s velocity and the Earth’s magnetic field is $65 . 7^{\circ}$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given Data

Length of shuttle (L) is 250.0 m.

Induced emf ( $ε$ ) is 40.0 V

Velocity of shuttle (v) is $7.8 \times 10^{3} m / s$

Magnetic field is $5.00 \times 10^{- 5} T$

Step 1: Define Electromagnetic Induction

In 1831, Michael Faraday made a discovery. He found that on placing a closed conductor in an everchanging magnetic field, an electromotive force is induced it. He named this phenomenon Electromagnetic induction.

Step 2: Evaluating the angle

As seen from the example, the induced emf is given as:

$ε = B L v \cos θ$

This allows us to solve to get the angle as:

$θ = \cos^{- 1} (\frac{ε}{B L v})$

Substituting Numerical values we get:

$θ = \cos^{- 1} (\frac{40 V}{5.00 \times 10^{- 5} T \times 250 m \times 7.8 \times 10^{3} m / s}) = 65 . 7^{\circ}$

Therefore, the required angle is: $65 . 7^{\circ}$ .

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College Physics (Urone)

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Short Answer

Step by Step Solution

Step 1: Given Data

Step 1: Define Electromagnetic Induction

Step 2: Evaluating the angle

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College Physics (Urone)

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Step by Step Solution

Step 1: Given Data

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Step 2: Evaluating the angle

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