 Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q22PE

Expert-verified Found in: Page 861 ### College Physics (Urone)

Book edition 1st Edition
Author(s) Paul Peter Urone
Pages 1272 pages
ISBN 9781938168000 # 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 $5.00×{10}^{-5}T$ field, while moving at $7.80×{10}^{3}m/s$ . 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 $65.{7}^{\circ }$.

See the step by step solution

## Step 1: Given Data

Length of shuttle (L) is 250.0 m.

Induced emf ($\epsilon$) is 40.0 V

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

Magnetic field is $5.00×{10}^{-5}T$

## 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:

$\epsilon =BLv\mathrm{cos}\theta$

This allows us to solve to get the angle as:

$\theta ={\mathrm{cos}}^{-1}\left(\frac{\epsilon }{BLv}\right)$

Substituting Numerical values we get:

$\theta ={\mathrm{cos}}^{-1}\left(\frac{40V}{5.00×{10}^{-5}T×250m×7.8×{10}^{3}m/s}\right)\phantom{\rule{0ex}{0ex}}=65.{7}^{\circ }$

Therefore, the required angle is: $65.{7}^{\circ }$. ### Want to see more solutions like these? 