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College Physics (Urone)
Found in: Page 861

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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 5.00×10-5 T field, while moving at 7.80×103m/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.

See the step by step solution

Step by Step Solution

Step 1: Given Data

Length of shuttle (L) is 250.0 m.

Induced emf (ε) is 40.0 V

Velocity of shuttle (v) is 7.8×103 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:


This allows us to solve to get the angle as:


Substituting Numerical values we get:

θ=cos-140 V5.00×10-5 T×250 m×7.8×103m/s =65.7

Therefore, the required angle is: 65.7.

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