To solve this problem we will use the Ampere-Maxwell law, which describes the magnetic fields that result from a transmitter wire or loop in electromagnetic surveys. According to Ampere-Maxwell law:

Where,
B= Magnetic Field
l = length
= Vacuum permeability
= Vacuum permittivity
Since the change in length (dl) by which the magnetic field moves is equivalent to the perimeter of the circumference and that the electric flow is the rate of change of the electric field by the area, we have to

Recall that the speed of light is equivalent to

Then replacing,


Our values are given as




Replacing we have,



Therefore the magnetic field around this circular area is 
I belive what your looking for is oxygen
Erosion, weathering, mechanical changes, chemical changes.
Really, any interaction can change the composition of a rock whether it be done by man or through nature.
I think its Mercury because it's the closest to the sun.
Potential energy U = mgh
Given h = 123 m,
mg = F = 780 N
Then
U = (123)(780)
= 95940
= 9.59 x 10^4