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Show that the drift speed of free electrons in a wire does not depend on the cross-sectional area of the wire.

Short Answer

Expert verified
Answer: No, the drift speed does not depend on the cross-sectional area. It is a function of the length of the wire (L) and the time (t) rather than its cross-sectional area (A).

Step by step solution

01

Recall the formula for drift speed

The drift speed (v_d) of free electrons in a wire can be expressed as: v_d = I / (n * A * q), where I is the current flowing through the wire, n is the number density of the free electrons (number of electrons per unit volume), A is the cross-sectional area of the wire, and q is the charge of an electron. Our goal is to show that the drift speed does not depend on the cross-sectional area (A) of the wire.
02

Express current in terms of charge and number of electrons

We know that the current flowing through a wire (I) can be expressed as the charge (Q) passing through the wire per unit time (t): I = Q / t, where Q = n * V * q. Here, V represents the volume of the wire through which the charge passes.
03

Relate the volume of the wire with its cross-sectional area

We can express the volume of the wire (V) as a product of its cross-sectional area (A) and length (L): V = A * L.
04

Substitute Q and V in the current formula

Now we substitute Q = n * V * q and V = A * L in the current formula. I = (n * A * L * q) / t.
05

Simplify the formula for drift speed

By rearranging the equation, we can find the drift speed (v_d) of the free electrons in the wire. v_d = I / (n * A * q) = (n * A * L * q) / (t * n * A * q) = L / t.
06

Conclude that drift speed does not depend on the cross-sectional area

From the above formula, we see that drift speed (v_d) is given by the ratio of length (L) to time (t), and there is no mention of the cross-sectional area (A) in this expression. Thus, we can conclude that the drift speed of free electrons in a wire does not depend on its cross-sectional area.

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Most popular questions from this chapter

Show that the power supplied to the circuit in the figure by the battery with internal resistance is maximum when the resistance of the resistor in the circuit, \(R\), is equal to \(R_{i}\). Determine the power supplied to \(R\). For practice, calculate the power dissipated by a \(12.0-\mathrm{V}\) battery with an internal resistance of \(2.00 \Omega\) when \(R=1.00 \Omega, R=2.00 \Omega,\) and \(R=3.00 \Omega\)

What would happen to the drift velocity of electrons in a wire if the resistance due to collisions between the electrons and the atoms in the crystal lattice of the metal disappeared?

A copper wire has a diameter \(d_{\mathrm{Cu}}=0.0500 \mathrm{~cm}\) is \(3.00 \mathrm{~m}\) long, and has a density of charge carriers of \(8.50 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\). As shown in the figure, the copper wire is attached to an equal length of aluminum wire with a diameter \(d_{\mathrm{A} \mathrm{I}}=0.0100 \mathrm{~cm}\) and density of charge carriers of \(6.02 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\). A current of 0.400 A flows through the copper wire. a) What is the ratio of the current densities in the two wires, \(J_{\mathrm{Cu}} / J_{\mathrm{Al}} ?\) b) What is the ratio of the drift velocities in the two wires, \(v_{\mathrm{d}-\mathrm{Cu}} / v_{\mathrm{d}-\mathrm{Al}} ?\)

If the current through a resistor is increased by a factor of \(2,\) how does this affect the power that is dissipated? a) It decreases by a factor of 4 . b) It increases by a factor of 2 . c) It decreases by a factor of 8 . d) It increases by a factor of 4 .

A thundercloud similar to the one described in Example 24.3 produces a lightning bolt that strikes a radio tower. If the lightning bolt transfers \(5.00 \mathrm{C}\) of charge in about \(0.100 \mathrm{~ms}\) and the potential remains constant at \(70.0 \mathrm{MV}\), find (a) the average current, (b) the average power, (c) the total energy, and (d) the effective resistance of the air during the lightning strike.

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