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
Something is reproducing.
Answer:
(a) A = 0.0800 m, λ = 20.9 m, f = 11.9 Hz
(b) 250 m/s
(c) 1250 N
(d) Positive x-direction
(e) 6.00 m/s
(f) 0.0365 m
Explanation:
(a) The standard form of the wave is:
y = A cos ((2πf) t ± (2π/λ) x)
where A is the amplitude, f is the frequency, and λ is the wavelength.
If the x term has a positive coefficient, the wave moves to the left.
If the x term has a negative coefficient, the wave moves to the right.
Therefore:
A = 0.0800 m
2π/λ = 0.300 m⁻¹
λ = 20.9 m
2πf = 75.0 rad/s
f = 11.9 Hz
(b) Velocity is wavelength times frequency.
v = λf
v = (20.9 m) (11.9 Hz)
v = 250 m/s
(c) The tension is:
T = v²ρ
where ρ is the mass per unit length.
T = (250 m/s)² (0.0200 kg/m)
T = 1250 N
(d) The x term has a negative coefficient, so the wave moves to the right (positive x-direction).
(e) The maximum transverse speed is Aω.
(0.0800 m) (75.0 rad/s)
6.00 m/s
(f) Plug in the values and find y.
y = (0.0800 m) cos((75.0 rad/s) (2.00 s) − (0.300 m⁻¹) (1.00 m))
y = 0.0365 m
The unmagnetized pieces of iron would be randomly pointing to directions, this is true because although influenced with the magnetic domain, the direction of the unmagnetized iron field of attraction is not uniform or does not have preferred direction.