Question

How large a current would a very long, straight wire have to carry so that the magnetic field 1.90 cm from the wire is equal to 1.00 G {comparable to the earth's northward-pointing magnetic field)? I = A If the wire is horizontal with the current running from east to west, at what locations would the magnetic field of the wire point in the same direction as the horizontal component of the earth's magnetic field? at all points directly north of the wire at all points directly south of the wire at all points directly below the wire at all points directly above the wire Repeat part (b) except the wire is vertical with the current going upward. at all points directly north of the wire at all points directly west of the wire

Answer 1

a) circuital Ampere's law states:
J = Integral over a closed path ( vector-B . vector-dr ) = M0. I
Where vector-dr is the infinitesimal displacement along the closed path
M0 = magnetic permeability of the vacuum
I = total current enclosed by the path

If we choose a circular path of radius 1.9x10⁻² cm with the centre at the straight wire and whose containing plane is perpendicular to the wire, then, from symmetry arguments:

The field B has constant modulus all along the path
the field B is everywhere tangent to that path
The above integral becomes:
J = 2.pi.R.B
Ampere's law thus gives:
2.pi.R.B = M0.I
I = 2.pi.R.B/M0
B = 1G but Gauss is not the SI unity for magnetic field, the SI unity is the Tesla
1G = 10^-4T
so,
M0=4.pi.10^-7 T.m/A
I = (2pi)*1.9x10^-2mx10^-4 T/[(4.pi)10^-7 T.m/A]
I = 9.5 A