Question

The Stanford Linear Accelerator accelerated a beam consisting of \(2.0 \cdot 10^{14}\) electrons per second through a potential difference of \(2.0 \cdot 10^{10} \mathrm{~V}\) a) Calculate the current in the beam. b) Calculate the power of the beam. c) Calculate the effective ohmic resistance of the accelerator.

Short answer

Question: In a linear accelerator, per second, 2.0 x 10^14 electrons get accelerated from rest, and potential difference of 2.0 x 10^10 V is applied. Calculate a) the current in the beam, b) the power of the beam, and c) the effective ohmic resistance of the accelerator. Answer: a) The current in the beam is 3.2 x 10^-5 A. b) The power of the beam is 6.4 x 10^5 W. c) The effective ohmic resistance of the accelerator is 6.25 x 10^14 Ω.

Step-by-step solution

a) Calculate the current in the beam

To find the current, we need the total charge accelerated per second. We will use the formula \(I = \frac{Q}{t}\), where \(I\) is the current, \(Q\) is the total charge, and \(t\) is the time. Since the time is one second, we can simplify the formula to \(I = Q\). We are given the number of electrons, \(n = 2.0 \cdot 10^{14}\), and we know the charge of one electron, \(e = 1.6 \cdot 10^{-19} \mathrm{C}\), so the total charge, \(Q\), in one second is: \(Q = n \cdot e = 2.0 \cdot 10^{14} \cdot 1.6 \cdot 10^{-19} \mathrm{C} = 3.2 \cdot 10^{-5} \mathrm{C}\) Therefore, the current in the beam is: \(I = Q = 3.2 \cdot 10^{-5} \mathrm{A}\).

b) Calculate the power of the beam

To calculate the power, we can use the formula \(P = IV\), where \(P\) is the power, \(I\) is the current, and \(V\) is the potential difference (voltage). We have the current from part a, \(I = 3.2 \cdot 10^{-5} \mathrm{A}\), and the potential difference from the exercise, \(V = 2.0 \cdot 10^{10} \mathrm{V}\), so we can calculate the power as: \(P = IV = (3.2 \cdot 10^{-5} \mathrm{A}) \cdot (2.0 \cdot 10^{10} \mathrm{V}) = 6.4 \cdot 10^{5} \mathrm{W}\).

c) Calculate the effective ohmic resistance of the accelerator

To find the effective ohmic resistance, we can use the formula \(R = \frac{V}{I}\), derived from Ohm's law. We have the potential difference, \(V = 2.0 \cdot 10^{10} \mathrm{V}\), and the current, \(I = 3.2 \cdot 10^{-5} \mathrm{A}\), so we can calculate the resistance as: \(R = \frac{V}{I} = \frac{2.0 \cdot 10^{10} \mathrm{V}}{3.2 \cdot 10^{-5} \mathrm{A}} = 6.25 \cdot 10^{14} \mathrm{\Omega}\). So the effective ohmic resistance of the accelerator is \(6.25 \cdot 10^{14} \mathrm{\Omega}\).