A250-turn solenoid with an area of 0.100m2 rotates at 40.orad/s in a 0.0250t uniform magnetic field that is perpendicular to the axis of rotation. calculate the instantaneous emf in the solenoid at the moment that the normal to its plane is at a 30.0° angle to the magnetic field. a. zero b. 12.5v c.21.6v d.25.0v e. 50.0v

B. 12.5 Volts

Explanation:

N = Number of turns of the solenoid = 250

B = magnitude of magnetic field = 0.0250 T

w = angular speed of rotation = 40 rad/s

A = Area = 0.100 m²

θ = angle between normal to the plane with magnetic field = 30.0°

E = Instantaneous EMF in the solenoid

Instantaneous EMF in the solenoid is given as

E = N B A w Sinθ

E = (250) (0.0250) (0.100) (40) Sin30

E = 12.5 Volts

answer: good job it's all correct !

explanation:

6 cos 34, 6 sin 34 trig