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In a current-carrying copper wire with cross section 𝜎 equals 4.0 millimeters squared, the drift velocity is 0.040 centimeters per second.
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Find the total current running through the wire.
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Use a value of 8.49 times 10 to the 28th inverse cubic meters for the number density of copper.
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In this wire, which has a cross-sectional area called 𝜎, there are electrons which move along with a drift velocity given in the problem statement.
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This drift velocity, along with the number density of electrons in this wire, will help us solve for the overall current that flows through the wire, which we can call 𝐼.
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As an equation, we can write that the current 𝐼 is equal to the number density of electrons in the wire times the magnitude of the charge on each one of those electrons multiplied by the cross-sectional area of the wire, given in our problem as 𝜎, times the drift velocity 𝑣 sub 𝑑.
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In the problem statement, we’re given the number density of electrons as well as their drift velocity and the cross-sectional area of our wire.
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We can approximate the magnitude of the charge on each charge carrier, the electrons, as 1.6 times 10 to the negative 19th coulombs.
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We’re now ready to plug in and solve for the current 𝐼 running through this copper wire.
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When we do, we’re careful to replace our cross-sectional area units with meters squared and to express the drift velocity of the electrons in meters per second.
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We do these conversions so the units of these terms are consistent with the rest of the terms in the expression.
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Looking over the units in each of our four terms, we see that the units of meters cubed in the denominator cancels out with meters squared times meters in the numerator.
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Our resulting units of this multiplication then will be coulombs per second.
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That is units of amperes.
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Since we’re solving for a current, this is a sign that we’re on the right track.
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To two significant figures, 𝐼 is 22 amperes.
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That’s the current running through this copper wire.