Answer:
Part (a) current in the wire is 1.144 A
Part (b) the drift velocity of electrons in the wire is 1.028 x 10⁻⁴ m/s
Explanation:
Given;
diameter d = 1.02 mm
current density J = 1.40×10⁶ A/m²
number of electron = 8.5×10²⁸ electrons
Part (a) Current in the wire
I = J×A
Where A is area of the wire;
[tex]A = \frac{\pi d^2}{4} \\\\A = \frac{\pi (1.02X10^{-3})^2}{4} = 8.1723 X10^{-7} m^2[/tex]
I = 1.40 x 10⁶ x 8.1723 x 10⁻⁷
I = 1.144 A
Part (b) the drift velocity of electrons in the wire
[tex]V = \frac{J}{nq} = \frac{1.4X10^6}{8.5X10^{28} X 1.602X10^{-19}} = 1.028 X10^{-4} m/s[/tex]
We were given the
diameter = 1.02 mm
current density = 1.40×10⁶ A/m²
number of electron = 8.5×10²⁸ electrons
We can use the formula:
I = J×A
where I is current, J is density and A is area.
A = π d²
4
= π (1.02ₓ 10⁻³)² = 8.1723 x 10⁻⁷
4
I = J×A
I = 1.40 x 10⁶ x 8.1723 x 10⁻⁷
I = 1.144 A
V = J/ nq
= 1.4 ₓ 10⁶ / (8.5ₓ 10²⁸ₓ 1.602ₓ 10⁻¹⁹)
= 1.028ₓ 10⁻⁴ m/s
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