Answer:
3.7 A
Explanation:
Parameters given:
Magnetic field strength, B = 5 * 10^(-5) T
Distance of magnetic field from wire, r = 1.5 cm = 0.015 m
The magnetic field, B, due to a current, I, flowing a wire is given as:
B = (μ₀*I) / 2πr
Where μ₀ = permeability of free space
To get the current, I, we make I the subject of the formula:
I = (2πr * B) / μ₀
I = (2 * 3.142 * 5 * 10^(-5)) / (1.25663706 × 10^(-6))
I = 3.7 A
Answer:
physic is branch of science in which math is its brother and will deal with the equation,law and evidence of natural phenomenon.
Answer:
a) 4.9*10^-6
b) 5.71*10^-15
Explanation:
Given
current, I = 3.8*10^-10A
Diameter, D = 2.5mm
n = 8.49*10^28
The equation for current density and speed drift is
J = I/A = (ne) Vd
A = πD²/4
A = π*0.0025²/4
A = π*6.25*10^-6/4
A = 4.9*10^-6
Now,
J = I/A
J = 3.8*10^-10/4.9*10^-6
J = 7.76*10^-5
Electron drift speed is
J = (ne) Vd
Vd = J/(ne)
Vd = 7.76*10^-5/(8.49*10^28)*(1.60*10^-19)
Vd = 7.76*10^-5/1.3584*10^10
Vd = 5.71*10^-15
Therefore, the current density and speed drift are 4.9*10^-6
And 5.71*10^-15 respectively
