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Chapter 18: Problem 3

In a region, \(10^{19} \alpha\)-particles and \(10^{19}\) protons move to the left, while \(10^{19}\) electrons move to the right per second. The current is (a) \(3.2\) A towards left (b) \(3.2\) A towards right (c) \(6.4\) A towards left (d) 6. 4 A towards right

Short Answer

Expert verified
The current is 6.4 A towards the left.

Step by step solution

01

Determine Charge Carried by Each Particle Type

To solve this problem, we first need to know the charge of each particle involved. The charge of 1 electron or 1 proton is approximately \(1.6 \times 10^{-19}\) C. An \(\alpha\)-particle consists of 2 protons and thus has a charge of \(2 \times 1.6 \times 10^{-19}\) C.
02

Calculate Total Charge of Alpha Particles

There are \(10^{19}\) \(\alpha\)-particles, each carrying a charge of \(3.2 \times 10^{-19}\) C (as calculated in Step 1). So, the total charge carried by the \(\alpha\)-particles is \(10^{19} \times 3.2 \times 10^{-19} = 3.2 \) C moving to the left per second.
03

Calculate Total Charge of Protons

Similarly, \(10^{19}\) protons each carry a charge of \(1.6 \times 10^{-19}\) C. Thus, the charge carried by all protons is \(10^{19} \times 1.6 \times 10^{-19} = 1.6 \) C moving to the left per second.
04

Calculate Total Charge of Electrons

\(10^{19}\) electrons each have a charge of \(-1.6 \times 10^{-19}\) C. Hence, the charge carried by all electrons is \(10^{19} \times -1.6 \times 10^{-19} = -1.6 \) C moving to the right per second.
05

Determine Net Current Direction and Magnitude

The net charge movement gives rise to the current. Charges moving left contribute positively and charges moving right contribute negatively. The total current can be calculated as follows:Current = Charge of protons + Charge of \(\alpha\)-particles - Charge of electrons \[ I = 1.6 \text{ C} + 3.2 \text{ C} - (-1.6 \text{ C}) = 6.4 \text{ C/s} \] Since positive charges (protons and \(\alpha\)-particles) dominate, the net current is to the left.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Alpha Particles
Alpha particles are a type of ionizing radiation often encountered in physics and chemistry. These particles are essentially helium nuclei, consisting of two protons and two neutrons. Due to the presence of two protons, alpha particles have a positive charge. Alpha particles are symbolized as \( \alpha \) and are represented as \( ^{4}_{2}\text{He}^{2+} \), reflecting their composition and charge.

Some key points to remember about alpha particles are:
  • They are positively charged due to the two protons they contain.
  • They have a relatively large mass compared to other subatomic particles such as electrons.
  • Alpha particles are not highly penetrating and can be stopped by a sheet of paper or human skin.
  • They play a crucial role in nuclear reactions and are typically found in processes like radioactive decay.
In current electricity, understanding the charge of alpha particles helps calculate the electric current as it contributes a significant amount of positive charge when moving in one direction.
The Charge of an Electron
Electrons are one of the fundamental particles and possess a negative charge. The value of this charge is approximately \(-1.6 \times 10^{-19}\) Coulombs. This is considered the basic unit of electric charge in many physics calculations. It's crucial to comprehend the nature of electrons because they are key players in determining the flow of electric current.

Let's delve into some important aspects of electron charge:
  • The negative charge of an electron balances the positive charge of a proton, thus maintaining electrical neutrality in atoms.
  • Electrons are much lighter compared to protons or alpha particles.
  • The movement of electrons from one atom to another is what constitutes electric current in most conductive materials.
In the context of the given exercise, electrons moving to the right contribute negatively to the total current. Calculating their effect on net charge flow is essential for determining the overall direction and magnitude of an electric current.
Direction of Electric Current
Electric current is defined as the flow of electric charge. Its direction is conventionally considered the direction of positive charge flow, even though electrons (which are negatively charged) are what's predominantly moving in conductors.

Here are key points on electric current direction:
  • Current direction is opposite to the flow of electrons in a wire.
  • Positive charges, like protons and alpha particles, when moving, set the conventional direction of the current.
  • In solving electrical problems, understanding this conventional flow is crucial to determining how current is quantified and its directional properties.
In the case of the exercise provided, the net current direction was determined by considering all moving charges. The total charge from protons and alpha particles moving left and electrons moving right resulted in a current of \(6.4\) Amperes towards the left. Grasping current direction is vital for analyzing electric circuits and predicting how components will behave under certain conditions.

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Most popular questions from this chapter

Four resistances \(40 \Omega, 60 \Omega, 90 \Omega\) and \(110 \Omega\) make the arms of a quadrilateral \(A B C D .\) Across \(A C\) is the battery circuit, the emf of the battery being \(4 \mathrm{~V}\) and internal resistance negligible. The potential difference across \(B D\) is (a) \(1 \mathrm{~V}\) (b) \(-1 \mathrm{~V}\) (c) \(-0.2 \mathrm{~V}\) (d) \(0.2 \mathrm{~V}\)

The resistance of a \(10 \mathrm{~m}\) long wire is \(10 \Omega\). Its length is increased by \(25 \%\) by stretching the wire uniformly. The resistance of wire will change to (approximately) (a) \(12.5 \Omega\) (b) \(14.5 \Omega\) (c) \(15.6 \Omega\) (d) \(16.6 \Omega\)

The number density of free electrons in a copper conductor estimated at \(8.5 \times 10^{28} \mathrm{~m}^{-3}\). How long does an electron take to drift from one end of a wire \(3.0 \mathrm{~m}\) long to its other end? The area of cross- section of the wire is \(2.0 \times 10^{-6} \mathrm{~m}^{2}\) and it is carrying a current of \(3.0 \mathrm{~A}\). (a) 6 h \(23 \mathrm{~min}\) (b) \(7 \mathrm{~h} 33 \mathrm{~min}\) (c) 7 h 43 min (d) \(6 \mathrm{~h} 53 \mathrm{~min}\)

The thermo emf \(E\) of a given thermocouple and 78 temperature \(\theta\) of the hot junction (with cold junction at \(0{ }^{\circ} \mathrm{C}\) ) are found to satisfy approximately the following relation. $$ E=a \theta+\frac{1}{2} b \theta^{2} $$ where \(E\) is in \(\mu \mathrm{V}, \theta\) in \({ }^{\circ} \mathrm{C}\) of the hot junction and and 79 \(a=14 \mu \mathrm{V}^{\circ} \mathrm{C}^{-1}\) and \(b=-0.04 \mu \mathrm{V}^{\circ} \mathrm{C}^{-2}\) At what temperature, the thermo emf in the given thermocouple is \(1.25 \mathrm{mV} ?\) (a) \(105^{\circ} \mathrm{C}\) (b) \(210^{\circ} \mathrm{C}\) (c) \(315^{\circ} \mathrm{C}\) (d) \(330^{\circ} \mathrm{C}\)

\(n\) identical cells, each of emf \(E\) and internal resistance \(r\), are connected in series a cell \(A\) is joined with reverse polarity. The potential difference across each cell, except \(A\) is (a) \(\frac{2 n E}{n-2}\) (b) \(\frac{(n-2) E}{n}\) (c) \(\frac{(n-1) E}{n}\) (d) \(\frac{2 E}{n}\)
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