A metal blade (length = 80 cm)spins at a constant rate of 10 radians/s about a pivot at one end.A uniform magnetic field of 2.0 mT is directed at an angle of 30 $\degree$ with the plane of the rotation.What is the magnitude of the potential difference between the two ends of the blade?
A)5.5 mV
B)6.4 mV
C)3.2 mV
D)11 mV
E)13 mV

Question

Quizplus · Accepted Answer

The potential difference (emf) induced in a rotating conductor in a magnetic field is given by the formula: emf = (1/2) * B * ω * L² * sin(θ), where B is the magnetic field strength, ω is the angular velocity, L is the length of the blade, and θ is the angle between the magnetic field and the plane of rotation. Substituting the given values: B = 2.0 mT = 2.0 x 10⁻³ T, ω = 10 rad/s, L = 0.8 m, and θ = 30°, we get emf = (1/2) * 2.0 x 10⁻³ * 10 * (0.8)² * sin(30°) = 3.2 mV.

A Metal Blade (Length = 80 Cm)spins at a Constant °\degree°

A Metal Blade (Length = 80 Cm)spins at a Constant $\degree$