A 500-turn circular coil with an area of 0.050 0 m² is mounted on a rotating frame that turns at a rate of 20.0 rad/s in the presence of a 0.050 0-T uniform magnetic field that is perpendicular to the axis of rotation.What is the instantaneous emf in the coil at the moment that the normal to its plane is at a 30.0 $\degree$ angle to the field? A)zero B)12.5 V C)21.6 V D)25.0 V

Question

Quizplus · Accepted Answer

The instantaneous emf in the coil is given by:emf = -N * (dΦ/dt)where N is the number of turns in the coil, Φ is the magnetic flux through the coil, and t is time.The magnetic flux through the coil is given by:Φ = B * A * cos(θ)where B is the magnetic field strength, A is the area of the coil, and θ is the angle between the normal to the plane of the coil and the magnetic field.In this problem, we have:N = 500 turnsA = 0.050 0 m^2B = 0.050 0 Tθ = 30.0°Plugging these values into the equation for magnetic flux, we get:Φ = (0.050 0 T) * (0.050 0 m^2) * cos(30.0°) = 2.16 * 10^-3 WbNow, we can plug this value into the equation for emf, along with the given angular velocity of 20.0 rad/s:emf = -500 turns * (-2.16 * 10^-3 Wb/s) = 12.5 VTherefore, the instantaneous emf in the coil at the moment that the normal to its plane is at a 30.0° angle to the field is 12.5 V.

A 500-Turn Circular Coil with an Area of 0 °\degree° Angle to the Field? A)zero B)12

A 500-Turn Circular Coil with an Area of 0 $\degree$ Angle to the Field?
A)zero
B)12