To solve this problem, we need to use the formula for current, I = nAvq, where n is the number density of free electrons, A is the cross-sectional area of the wire, v is the drift speed, and q is the charge of an electron.

First, we need to find the current in the wire. We can use Ohm's law, V = IR, where V is the potential difference, I is the current, and R is the resistance. Rearranging for I gives us I = V/R = 6.0V / 15Ω = 0.4A.

Next, we need to find the cross-sectional area of the wire. The area of a circle is given by A = πr², where r is the radius. The diameter of the wire is 3.0mm, so the radius is 1.5mm or 1.5 x 10^-3 m. Substituting this into the formula gives us A = π(1.5 x 10^-3 m)² = 7.07 x 10^-6 m².

Finally, we can substitute these values into the formula for current. Rearranging for v gives us v = I / (nAq). The charge of an electron is 1.6 x 10^-19 C. Substituting all the values gives us v = 0.4A / (9.0 x 10^28 m^-3 * 7.07 x 10^-6 m² * 1.6 x 10^-19 C) = 3.9 x 10^-6 m/s.

So, the average drift speed of the free electrons in the wire is 3.9 x 10^-6 m/s, which corresponds to option B.

Question

To solve this problem, we need to use the formula for current, I = nAvq, where n is the number density of free electrons, A is the cross-sectional area of the wire, v is the drift speed, and q is the charge of an electron.

First, we need to find the current in the wire. We can use Ohm's law, V = IR, where V is the potential difference, I is the current, and R is the resistance. Rearranging for I gives us I = V/R = 6.0V / 15Ω = 0.4A.

Next, we need to find the cross-sectional area of the wire. The area of a circle is given by A = πr², where r is the radius. The diameter of the wire is 3.0mm, so the radius is 1.5mm or 1.5 x 10^-3 m. Substituting this into the formula gives us A = π(1.5 x 10^-3 m)² = 7.07 x 10^-6 m².

Finally, we can substitute these values into the formula for current. Rearranging for v gives us v = I / (nAq). The charge of an electron is 1.6 x 10^-19 C. Substituting all the values gives us v = 0.4A / (9.0 x 10^28 m^-3 * 7.07 x 10^-6 m² * 1.6 x 10^-19 C) = 3.9 x 10^-6 m/s.

So, the average drift speed of the free electrons in the wire is 3.9 x 10^-6 m/s, which corresponds to option B.

Knowee AI · Accepted Answer

To solve this problem, we need to use the formula for current, I = nAvq, where n is the number density of free electrons, A is the cross-sectional area of the wire, v is the drift speed, and q is the charge of an electron.

First, we need to find the current in the wire. We can use Ohm's law, V = IR, where V is the potential difference, I is the current, and R is the resistance. Rearranging for I gives us I = V/R = 6.0V / 15Ω = 0.4A.

Next, we need to find the cross-sectional area of the wire. The area of a circle is given by A = πr², where r is the radius. The diameter of the wire is 3.0mm, so the radius is 1.5mm or 1.5 x 10^-3 m. Substituting this into the formula gives us A = π(1.5 x 10^-3 m)² = 7.07 x 10^-6 m².

Finally, we can substitute these values into the formula for current. Rearranging for v gives us v = I / (nAq). The charge of an electron is 1.6 x 10^-19 C. Substituting all the values gives us v = 0.4A / (9.0 x 10^28 m^-3 * 7.07 x 10^-6 m² * 1.6 x 10^-19 C) = 3.9 x 10^-6 m/s.

So, the average drift speed of the free electrons in the wire is 3.9 x 10^-6 m/s, which corresponds to option B.

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