To solve this problem it is necessary to apply the concepts related to the magnetic dipole moment in terms of the current and the surface area, as well as the current density, as a function of the current over the area.
Part A) By definition we know that magnetic dipole moment is
Where,
I = Current
S = Area
Replacing with our values we have that,
Re-arrange to find I,
Part B) To find the Current density we need to find the cross sectional area of the Wire:
Finally the current density is simply J
PART C) Finally to make the comparison with the given values we have to cross-sectional area would be
Therefore the current density would be
Comparing the two values we can see that the 2mm wire has a higher current density.
Answer:the sample of solid has less energy than the sample of gas
Explanation:
Apex
What’s the question though
If that's the case, then
50 units = 0.55 x the input energy
Divide each side by 0.55 :
50 units/0.55 = the input energy =
<em> 90 and 10/11 units</em>
Answer:
Let N0 be the initial atoms of Be11
N0 / 2^1 = N0 / 2 Number of Be11 after 1 half-life
N0 / 2^2 = N0/ 4 Number of Be11 after 2 half-lives
30/13.81 = 2.17 half lives
N0 / 2^2.17 = N0 / 4.51 = .222 N0 or 22.2 % of atoms remain