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Chapter 23: Q22PE (page 861)

In the August 1992space shuttle flight, only 250 mof the conducting tether considered in Example 23.2could be let out. A 40.0 Vmotional emf was generated in the Earth's 5.00×10-5Tfield, while moving at7.80×103m/s. What was the angle between the shuttle’s velocity and the Earth’s field, assuming the conductor was perpendicular to the field?

Short Answer

Expert verified

The angle between the shuttle’s velocity and the Earth’s magnetic field is 65.7∘.

Step by step solution

01

Given Data

Length of shuttle (L) is 250.0 m.

Induced emf (ε) is 40.0 V

Velocity of shuttle (v) is 7.8×103m/s

Magnetic field is5.00×10-5T

02

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.

03

Evaluating the angle

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

ε=BLvcosθ

This allows us to solve to get the angle as:

θ=cos-1εBLv

Substituting Numerical values we get:

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

Therefore, the required angle is: 65.7∘.

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