Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 25: Problem 73

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

Expert verified
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

01

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: \(I = Q = 3.2 \cdot 10^{-5} \mathrm{A}\).
02

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: \(P = IV = (3.2 \cdot 10^{-5} \mathrm{A}) \cdot (2.0 \cdot 10^{10} \mathrm{V}) = 6.4 \cdot 10^{5} \mathrm{W}\).
03

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 \(6.25 \cdot 10^{14} \mathrm{\Omega}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

What is the current density in an aluminum wire having a radius of \(1.00 \mathrm{~mm}\) and carrying a current of \(1.00 \mathrm{~mA}\) ? What is the drift speed of the electrons carrying this current? The density of aluminum is \(2.70 \cdot 10^{3} \mathrm{~kg} / \mathrm{m}^{3},\) and 1 mole of aluminum has a mass of \(26.98 \mathrm{~g}\). There is one conduction electron per atom in aluminum.

Two conductors of the same length and radius are connected to the same emf device. If the resistance of one is twice that of the other, to which conductor is more power delivered?

A certain brand of hot dog cooker applies a potential difference of \(120 \mathrm{~V}\) to opposite ends of the hot dog and cooks it by means of the heat produced. If \(48 \mathrm{~kJ}\) is needed to cook each hot dog, what current is needed to cook three hot dogs simultaneously in \(2.0 \mathrm{~min}\) ? Assume a parallel connection.

A copper wire that is \(1 \mathrm{~m}\) long and has a radius of \(0.5 \mathrm{~mm}\) is stretched to a length of \(2 \mathrm{~m}\). What is the fractional change in resistance, \(\Delta R / R,\) as the wire is stretched? What is \(\Delta R / R\) for a wire of the same initial dimensions made out of aluminum?

A battery has a potential difference of \(14.50 \mathrm{~V}\) when it is not connected in a circuit. When a \(17.91-\Omega\) resistor is connected across the battery, the potential difference of the battery drops to \(12.68 \mathrm{~V}\). What is the internal resistance of the battery?
See all solutions

Recommended explanations on Physics Textbooks

Magnetism

Read Explanation

Space Physics

Read Explanation

Electric Charge Field and Potential

Read Explanation

Circular Motion and Gravitation

Read Explanation

Fields in Physics

Read Explanation

Fundamentals of Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free