Question

Find the magnitude and direction of the weakest magnetic field required to levitate a wire of length 4.90 cm and mass 1.88 g, carrying a current of 7.52 A.

Answer 1

So in other words, you want to find the magnetic force required to counteract the force of gravity pulling the wire down.


The first step is to find what the magnitude of this force is. We can do this by calculating the force due to gravity acting on the wire.

F = mg = (0.00188) * (9.8) = 0.0184 N


Now we need to compute the magnetic field required to produce a magnetic force of this magnitude. In order to do this, we must use Lorentz's Force Law.

F = q (V x B) can be rewritten as: F = I (L x B)

I = current in wire = 7.52 A
L = length of wire = 0.049 m
B = magnetic field strength = ? <-- what we want to find
F = magnetic force = mg = 0.0184 N

L x B magnitude would be L*B*sin(angle), but in this case, the angle is 90-degrees, so sine goes to 1.

B = F / I*L = (0.0184) / [(7.52) * (0.049)] = 0.05 T

So you would need a field of 0.05 Tesla to support the wire.


Now for the direction. We used the simplification that the angle was 90-degrees. Therefore, we need to now take that into account of how the magnetic field needs to be orientated. The direction is shown in the drawing below.