A potential difference of \(V=0.500 \mathrm{~V}\) is applied across a block of silicon with resistivity \(8.70 \cdot 10^{-4} \Omega \mathrm{m}\). As indicated in the figure, the dimensions of the silicon block are width \(a=2.00 \mathrm{~mm}\) and length \(L=15.0 \mathrm{~cm} .\) The resistance of the silicon block is \(50.0 \Omega\), and the density of charge carriers is \(1.23 \cdot 10^{23} \mathrm{~m}^{-3}\) Assume that the current density in the block is uniform and that current flows in silicon according to Ohm's Law. The total length of 0.500 -mm-diameter copper wire in the circuit is \(75.0 \mathrm{~cm},\) and the resistivity of copper is \(1.69 \cdot 10^{-8} \Omega \mathrm{m}\) a) What is the resistance, \(R_{w}\) of the copper wire? b) What are the direction and the magnitude of the electric current, \(i\), in the block? c) What is the thickness, \(b\), of the block? d) On average, how long does it take an electron to pass from one end of the block to the other? \(?\) e) How much power, \(P\), is dissipated by the block? f) In what form of energy does this dissipated power appear?

Step by step solution

01

a) Calculating the resistance of the copper wire

First, we will find the resistance of the copper wire using the resistivity formula: \(R_w = \frac{\rho_w \cdot L_w}{A_w}\) We know that the resistivity of copper (\(\rho_w\)) is \(1.69 \cdot 10^{-8} \Omega m\), and the total length of the copper wire (\(L_w\)) is \(75.0 cm = 0.75 m\). The wire's cross-sectional area (\(A_w\)) can be found using the formula for the area of a circle, as the wire is cylindrical. The diameter of the wire is \(0.500 mm = 5.00 \cdot 10^{-4} m\), so its radius (\(r_w\)) is half of that: \(r_w = 2.50 \cdot 10^{-4} m\) Now we can find the area: \(A_w = \pi r_w^2 = \pi (2.50 \cdot 10^{-4})^2\) Plug all the values into the resistivity formula to find the wire resistance: \(R_w = \frac{1.69 \cdot 10^{-8} \cdot 0.75}{\pi (2.50 \cdot 10^{-4})^2} \approx 0.000169 \Omega\) So the resistance of the copper wire is approximately \(0.169 \times 10^{-3} \Omega\).

02

b) Calculating the direction and magnitude of electric current in the block

We are given that the resistance of the silicon block is \(50.0 \Omega\) and the potential difference (V) across the block is \(0.5 V\). According to Ohm's Law, the electric current (i) can be found using the formula: \(i = \frac{V}{R_{total}}\) The total resistance (\(R_{total}\)) includes both the resistance of the silicon block and the resistance of the copper wire: \(R_{total} = 50.0 + 0.000169 \approx 50.0 \Omega\) By Ohm's Law, the electric current (i) is: \(i = \frac{0.5}{50} = 0.01 A\) So the magnitude of the electric current in the block is \(0.01 A\). The direction of the current in the block is the same as the direction of current flow from the positive terminal of the voltage source to the negative terminal.

03

c) Calculating the thickness of the silicon block

Now we will find the thickness (b) of the silicon block using the resistivity formula. We have the block's resistivity, width, and length, so we can rewrite the formula in terms of its thickness: \(R_{block} = \frac{\rho_{block} \cdot L}{A_{block}}\) Now, \(A_{block} = a \cdot b\). Therefore, \(R_{block} = \frac{\rho_{block} \cdot L}{a \cdot b}\) Plugging in the given values, we have: \(50 \Omega = \frac{8.7 \times 10^{-4} \Omega m \cdot 0.15 m}{0.002 m \cdot b}\) Rearrange the equation and solve for b: \(b = \frac{8.7 \times 10^{-4} \Omega m \cdot 0.15 m}{50 \Omega \cdot 0.002 m} \approx 1.3 \times 10^{-3} m\) So the thickness of the silicon block is approximately \(1.3 mm\).

04

d) Calculating the average time an electron takes to pass through the block

We first need to find the drift velocity (\(v_d\)) of the electrons using the formula: \(v_d = \frac{I}{n \cdot q \cdot A_{block}}\) Where n is the density of charge carriers (\(1.23 \cdot 10^{23} m^{-3}\)), and q is the charge of an electron (\(1.6 \times 10^{-19} C\)). We already know the cross-sectional area of the block, and the current in the block is \(0.01 A\). So, \(v_d = \frac{0.01}{1.23 \cdot 10^{23} \cdot 1.6 \times 10^{-19} \cdot (2 \times 10^{-3} \cdot 1.3 \times 10^{-3})} \approx 2.4 \times 10^{-5} m/s\) Now we can find the average time (t) an electron takes to pass through the block using the formula: \(t = \frac{L}{v_d}\) \(t = \frac{0.15}{2.4 \times 10^{-5}} \approx 6.25 \times 10^3 s\) So on average, it takes an electron approximately \(6.25 \times 10^3\) seconds to pass from one end of the block to the other.

05

e) Calculating the power dissipated by the silicon block

Now we will find the power (P) dissipated by the silicon block using the formula: \(P = I^2 \cdot R_{block}\) Plug in the current (\(0.01 A\)) and the resistance of the silicon block (\(50 \Omega\)): \(P = (0.01)^2 \cdot 50 \approx 5 \times 10^{-3} W\) So, the power dissipated by the silicon block is approximately \(5 mW\).

06

f) Identifying the form of energy from dissipated power

The energy dissipated as power in the silicon block is in the form of heat. This is because, when the block resists the flow of the electric current, the kinetic energy of moving electrons gets converted into heat energy. This increases the temperature of the block and causes it to dissipate heat to its surroundings.

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