What is the binding energy of Helium atom if its mass is 4.002603 u?
1 answer:
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Answer:
B = 38.2μT
Explanation:
By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:
(1)
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
r: distance from the center of the cylinder, in which B is calculated
Ir: current for the distance r
In this case, you first calculate the current Ir, by using the following relation:
J: current density
Ar: cross sectional area for r in the hollow cylinder
Ar is given by
The current density is given by the total area and the total current:
R2: outer radius = 26mm = 26*10^-3 m
R1: inner radius = 5 mm = 5*10^-3 m
IT: total current = 4 A
Then, the current in the wire for a distance r is:
(2)
You replace the last result of equation (2) into the equation (1):
Finally. you replace the values of all parameters:
hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT