Question

In the August 1992 space shuttle flight, with 250 m of conducting tether, generating a 40.0 V motional emf in Earth’s 5.00×10^−5T field at 7.80×10^3m/s, the angle θ between shuttle’s velocity and Earth’s field (assuming the conductor is perpendicular to the field) is:

a) 15.3°
b) 30.7°
c) 45.0°
d) 60.4°

Answer 1

The angle between the shuttle's velocity and Earth's magnetic field can be solved using the motional emf formula. Substituting the given values into the rearranged formula, we find that the sine of the angle is 0.516129, which corresponds to an angle of approximately 31 degrees.

Step-by-step explanation:

To solve for the angle between the shuttle's velocity and Earth's magnetic field, we can use the formula for the motional electromotive force (emf), which is given by № = Bvl sin θ, where 'emf' is the motional emf, 'B' is the magnetic field strength, 'l' is the length of the conductor, 'v' is the velocity of the conductor, and θ is the angle between the velocity and the magnetic field. We have been given the following values:

• Motional emf (emf) = 40.0 V
• Magnetic field strength (B) = 5.00 × 10⁻⁵ T
• Length of the conductor (l) = 250 m
• Velocity of the conductor (v) = 7.80 × 10³ m/s

The formula for the motional emf can be rearranged to solve for the sine of angle θ:

sin θ = № / (Bvl)

Substituting the given values:

sin θ = 40.0 V / (5.00 × 10⁻⁵ T × 250 m × 7.80 × 10³ m/s)

sin θ = 0.516129

Therefore, the angle θ = sin⁻¹(0.516129) which is approximately 31 degrees. Hence, the correct answer for the angle between the shuttle's velocity and Earth's magnetic field is 30.7° (Option B).