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Zorluk: OrtaElectric Current and Resistance

A uniform copper wire with a cross-sectional area of 2.0×10−6 m22.0 \times 10^{-6}\,\text{m}^2 carries a steady electric current of 4.0 A4.0\,\text{A}. Given that the free electron density of copper is 2.5×1028 m−32.5 \times 10^{28}\,\text{m}^{-3} and the elementary charge is 1.6×10−19 C1.6 \times 10^{-19}\,\text{C}, calculate the drift velocity of the electrons in the wire in m/s\text{m/s}.

Cevap: 0.0005 m/s

Cevap

The drift velocity of the electrons in the wire is 0.0005 m/s0.0005\,\text{m/s} (or 5.0×10−4 m/s5.0 \times 10^{-4}\,\text{m/s}).
Applying the formula I=nAevdI = n A e v_d and solving for drift velocity yields vd=InAe=4.0(2.5×1028)(2.0×10−6)(1.6×10−19)=4.08000=0.0005 m/sv_d = \frac{I}{n A e} = \frac{4.0}{(2.5 \times 10^{28})(2.0 \times 10^{-6})(1.6 \times 10^{-19})} = \frac{4.0}{8000} = 0.0005\,\text{m/s}.

Adım Adım Çözüm

1
State the formula relating electric current to microscopic drift velocity.
I=nAevdI = n A e v_d
Electric current is the net charge flowing per unit time across a conductor's cross-section.
2
Isolate the drift velocity vdv_d as the target variable.
vd=InAev_d = \frac{I}{n A e}
Algebraic rearrangement places all known quantities on the right-hand side.
3
Substitute the numerical values into the equation.
vd=4.0(2.5×1028)(2.0×10−6)(1.6×10−19)v_d = \frac{4.0}{(2.5 \times 10^{28})(2.0 \times 10^{-6})(1.6 \times 10^{-19})}
Inserting I=4.0 AI = 4.0\,\text{A}, n=2.5×1028 m−3n = 2.5 \times 10^{28}\,\text{m}^{-3}, A=2.0×10−6 m2A = 2.0 \times 10^{-6}\,\text{m}^2, and e=1.6×10−19 Ce = 1.6 \times 10^{-19}\,\text{C}.
4
Evaluate the arithmetic computation.
vd=0.0005 m/sv_d = 0.0005\,\text{m/s}
Dividing 4.04.0 by 80008000 yields 0.0005 m/s0.0005\,\text{m/s}.

Anahtar Kavram

Relationship between electric current, charge carrier density, and drift velocity
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Konular

Electric Current and Resistance

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