To solve this problem, we can break it down step by step using formulas from electromagnetism and current conduction. The parameters given in the problem are:

• Cross-sectional area $A = 3.97 \, \text{cm}^2 = 3.97 \times 10^{-4} \, \text{m}^2$
• Front-to-rear length $L = 20.8 \, \text{cm} = 0.208 \, \text{m}$
• Resistance $R = 1150 \, \Omega$
• Number density of conduction electrons $n = 5.78 \times 10^{22} \, \text{electrons/m}^3$
• Potential difference $V = 44.9 \, \text{V}$

We'll use these to find:

(a) The current in the block

Ohm's law relates the current $I$ through a material to the potential difference $V$ and the resistance $R$:

$I = \frac{V}{R}$

Substitute the given values:

$I = \frac{44.9 \, \text{V}}{1150 \, \Omega} = 0.03904 \, \text{A}$

Thus, the current in the block is $I = 0.03904 \, \text{A}$ or 39.04 mA.

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(b) The magnitude of the current density

The current density $J$ is related to the current $I$ and the cross-sectional area $A$ by the formula:

$J = \frac{I}{A}$

Using the current from part (a) and the area:

$J = \frac{0.03904 \, \text{A}}{3.97 \times 10^{-4} \, \text{m}^2} = 98.37 \, \text{A/m}^2$

Thus, the current density is $J = 98.37 \, \text{A/m}^2$.

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(c) The drift velocity of the conduction electrons

The drift velocity $v_d$ can be found using the relation:

$v_d = \frac{J}{n e}$

Where:

• $J = 98.37 \, \text{A/m}^2$ (from part (b))
• $n = 5.78 \times 10^{22} \, \text{electrons/m}^3$
• $e = 1.60 \times 10^{-19} \, \text{C}$ (the charge of an electron)

Substitute the values:

$v_d = \frac{98.37 \, \text{A/m}^2}{(5.78 \times 10^{22} \, \text{electrons/m}^3)(1.60 \times 10^{-19} \, \text{C})}$

$v_d = 1.06 \times 10^{-7} \, \text{m/s}$

Thus, the drift velocity is $v_d = 1.06 \times 10^{-7} \, \text{m/s}$.

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(d) The magnitude of the electric field in the block

The electric field $E$ can be found using Ohm's law in its form for electric fields and potentials:

$E = \frac{V}{L}$

Substitute the given values:

$E = \frac{44.9 \, \text{V}}{0.208 \, \text{m}} = 215.87 \, \text{V/m}$

Thus, the magnitude of the electric field in the block is $E = 215.87 \, \text{V/m}$.

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Summary of results:

• (a) Current in the block: $0.03904 \, \text{A}$ or 39.04 mA
• (b) Current density: $98.37 \, \text{A/m}^2$
• (c) Drift velocity of the conduction electrons: $1.06 \times 10^{-7} \, \text{m/s}$
• (d) Magnitude of the electric field: $215.87 \, \text{V/m}$

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Would you like further details on any part of this explanation?

Here are five related questions to expand your understanding:

1. How does the resistivity of the material affect the current in the block?
2. What factors influence the drift velocity of electrons in a conductor?
3. How would doubling the length of the block affect the electric field?
4. Can you calculate the power dissipated in the block?
5. What is the relationship between current density and resistivity in a material?

Tip: The drift velocity of electrons is typically very small, but the large number of electrons in a conductor allows significant currents to flow.