# Magnetic flux and emf help?

ghostbuster25
The Earth’s magnetic field at a certain location in the UK has a magnitude of 48 μT and is directed at 66° below the horizontal.

a)Determine the magnitude of the flux of the Earth’s magnetic field through a
wedding ring of diameter 2.2 cm when the ring is held in a horizontal plane.

b)What is the magnitude of the flux change when the ring is flipped through 180° about a horizontal axis?

c)Use Faraday’s law to calculate the EMF induced in the ring when it is flipped
through 180° in 0.2 s, assuming the flux changes uniformly during this time.

can anyone help me please i am completely stuck and don't know where to start :(

Mindscrape
You would actually probably be better off posting this in introductory physics, but that's okay, let me give you a hand anyway.

a) Remember that the equation for flux is $$\int \int \mathbf{B}\cdot d\mathbf{a}=\int \int B cos(\phi) da$$

c) Remember that EMF is $$\frac{d \Phi}{dt}$$

cerium
Hi ghostbuster remember the B dot A cross product rule that helped me and the angle is below thw horizontal so work out the angle you need

billmcalpine7
Not all that astonishingly, I have the same homework as ghostbuster. it's getting pretty late now though, but here are my tenuous attempts at the three questions:

a)

Magnetic flux = B dot A; this is the dot product of the B vector and the area vector, A

B = 48 x10^-6 T at 34 degrees from the normal vector to the the area

The A vector has magnitude = the area, and is directed perpendicular to the surface.
Mag A = (2.2 x 10^-2 m)^2

Therefore
flux = (48 x 10^-6 T) (Pi) [(2.2 x 10^-2 m)^2] (cos 34)
= 2.75 x 10^-6 T m ^2

(b)

The A vector changes direction

Δ flux = B A cos (156) – BA cos (34)
= ( -0.9135 - 0.8290) x 10^-8
= 1.7425 x 10^-8 T m2

[c]