A metal blade spins at a constant rate of 5.0 revolutions per second about a pivot through one end of the blade.This rotation occurs in a region where the component of the earth's magnetic field perpendicular to the blade is 30 $\mu$T.If the blade is 60 cm in length,what is the magnitude of the potential difference between its ends?

Question

Quizplus · Accepted Answer

The potential difference induced across the length of the spinning blade is given by:
V = B * L * ω
where B is the magnetic field, L is the length of the blade, and ω is the angular frequency of rotation.

Here, B = 30 μT = 30 × 10^-6 T, L = 60 cm = 0.6 m, and ω = 2πf = 2π × 5.0 = 31.4 rad/s.

Substituting these values, we get:
V = (30 × 10^-6) * 0.6 * 31.4 = 0.01776 mV = 0.17 mV (approx)

Therefore, the magnitude of potential difference between the ends of the blade is 0.17 mV, which is closest to option C.

A Metal Blade Spins at a Constant Rate of 5 μ\muμ

A Metal Blade Spins at a Constant Rate of 5 $\mu$