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Q22PE

Expert-verifiedFound in: Page 861

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 $5.00\times {10}^{-5}T$ ****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 $65.{7}^{\circ}$.

Length of shuttle (L) is 250.0 m.

Induced emf ($\epsilon $) is 40.0 V

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

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

**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.**

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:

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

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

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