Question

A loop has a radius of 15 centimeters and it is initially in the plane of your screen. A constant magnetic field is pointing out of the screen and has a value of 1T. The loop then rotates by 45º along a vertical axis (so it's partially sticking out of your screen). It takes 4.8 seconds to rotate. Determine the electromotive force (the voltage) that will be generated in the process (express your answer in mV)

Answer 1

Radius of the loop, r = 15 cm = 0.15 m

Area of the loop, A = π* r^{​​​​​​2}​​​ = 0.0707 m^{​​​​​​2} . Direction of the surface area is perpendicular to the plane of the loop, which is pointing out of the page initially. And then rotates 45°

Magnetic field, B= 1.0 T

Magnetic flux, = B*A* cos, where is the angle between the loop surface and magnetic field.

Initial flux ₁ = B*A*cos(0°) = B*A

Final flux ₂ = B*A*cos(45°)

Change of flux = ₂ – ₁ = B*A* cos(45°) – B*A = 1×0.0707× cos(45°) –1×0.0707 = – 0.0207 Wb

Time taken t = 4.8 s

Electromotive force (emf) = – = –(–0.0207/4.8) = 0.00431 V = 4.31 mV