A long, straight, horizontal wire carries current toward the east. A proton moves toward the east alongside and just south of th
1 answer:
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As per impulse momentum theorem we know that
now here we will have
t = 1.30 ms
m = 0.144 kg
now we need to find final speed using above formula
so final speed is given as above
I am assuming you know the relation obtained between slit width, distance of screen from slits, distance of interference pattern obtained on the screen from the center and the wavelength of monochromatic light used in Young's Double Slit experiment.
λ =
λ ~ 1.97 ×10⁻⁷m
λ =
λ ~ 1.97 ×10⁻⁷m
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