# Magnetic Flux and Induced EMF at South Pole

1. Apr 25, 2013

### m00nbeam360

1. The problem statement, all variables and given/known data
The magnetic field strength at the South Pole is around 66μT, and magnetic field lines point out of the earth perpendicular to the surface. The sled is a metal square of side 0.95 m (with non-conducting wooden slats in between).

What is the magnetic flux through the sled sitting flat on the ground?

Imagine that you accidentally flip the sled from horizontal to vertical in 0.65 s. What is the induced emf in the sled as a result of that motion?

2. Relevant equations

B = 0.66 x 10^-3 T + z direction
A = (0.95m)^2 = 0.9025 m
θ = 90° (?)

Phi = B * nA = BAcosθ = BnA

ε = -dPhi/dt

3. The attempt at a solution

I tried using the flux equation with cos(90), but since that's 0, would the flux really be 0?? Or is it just B*A?

Same problem with induced emf, but how do I calculate the derivative?

Thanks!

2. Apr 25, 2013

### Staff: Mentor

The angle between the magnetic field and the normal direction of the area is 0, not 90°.
B*A, right.
As we don't have details about the process, just consider the flux before and after the flip, and how quick it happens.

3. Apr 25, 2013

### barryj

If the sled is flat on the ground, and the field lines perpendicular to the surface, then the flux will simply be the area of the sled X B. As you point out, ε = -dPhi/dt . When the sled is vertical, there is no field lines passing through the sled. It takes .65 sec to flip. This enough hints??

4. Apr 26, 2013

### m00nbeam360

Thanks for the help! So is it really just (5.957 * 10^-4 Wb/m^2)/ (0.65 s)? Seems too easy.

5. Apr 26, 2013

### Staff: Mentor

Don't forget the area, otherwise the units do not match.