/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Ion Source An ion source is prod... [FREE SOLUTION] | 91Ó°ÊÓ

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Ion Source An ion source is producing ions of \({ }^{6} \mathrm{Li}\) (mass = \(6.0 \mathrm{u}\) ), each with a charge of \(+e\). The ions are accelerated by a potential difference of \(10 \mathrm{kV}\) and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude \(|\vec{B}|=\) 1.2 T. Calculate the strength of the smallest electric field, to be set up over the same region, that will allow the \({ }^{6} \mathrm{Li}\) ions to pass through undeflected.

Short Answer

Expert verified
The smallest electric field needed is 1.2 kV/cm.

Step by step solution

01

Understand the Problem

Determine the conditions under which the Li ions pass through without deflection in a region with a magnetic and electric field. This means the forces from both fields must cancel each other.
02

Calculate the kinetic energy

The kinetic energy (\(K.E.\)) of the ions can be found using the potential difference (\(V\)): \(K.E. = eV\). Here \(e = 1.602 \times 10^{-19}\)C and \(V = 10\) kV.
03

Calculate the speed of the ions

Relate the kinetic energy to the ions' speed (\(v\)) with the mass (\(m\)) of \({ }^{6} \text{Li}\): \(K.E. = \frac{1}{2} mv^2\). Solve for \(v\).
04

Set up the force balance equation

For the ions to be undeflected, the electric force must balance the magnetic force. That is \(eE = evB\). Solve for \(E\).
05

Substitute and calculate

Use the values from previous steps to find the electric field strength, \(E\).

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Key Concepts

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

Electric Field
An electric field (\textbf{E}) is a region around a charged particle where a force would be experienced by other charges. In the context of our problem, the electric field needed is to counteract the deflection caused by the magnetic field, ensuring the ion travels straight. When the ion enters the electric field, it feels a force proportional to the charge of the ion times the electric field strength. Mathematically, this is expressed as:
\(F_E = eE\),
where \(e\) is the charge of the ion (for \(^6 Li^\text{+}\), this is \(+e\)). The direction of this force is determined by the direction of the electric field and the sign of the charge. This force ensures that the ion remains undeviated by balancing out the magnetic force.
Magnetic Field
Magnetic fields (\textbf{B}) exert a force on moving charges, such as ions. When a charged particle moves through a magnetic field perpendicular to its velocity, it experiences a force termed as the Lorentz force. This force acts perpendicular to both the magnetic field direction and the velocity of the particle and is given by:
\(F_B = evB\),
where \(v\) is the velocity of the ion and \(B\) is the magnetic field strength. The magnetic force would normally cause the ion to move in a curved path, but if an electric field is properly applied, this effect can be counteracted.
Potential Difference
Potential difference (\textbf{V}) is the difference in electric potential between two points. It tells us how much work is done in moving a charge between those two points. In the exercise, the ions are accelerated by a potential difference of \(10 \text{kV}\). This energy provided to the ions is converted into kinetic energy, calculated using:
\(K.E. = eV\),
where \(e\) is the elementary charge \(1.602 \times 10^{-19}\text{C}\) and \(V\) is the provided potential difference. This kinetic energy gained by the ions plays a crucial role in determining their speed.
Kinetic Energy
Kinetic energy (\textbf{K.E.}) relates to the energy that a body possesses due to its motion. When the ions are accelerated by the potential difference, their kinetic energy can be found using the equation:
\(K.E. = \frac{1}{2} mv^2\),
where \(m\) is the mass of the ion and \(v\) is its velocity. First, we get the kinetic energy from the potential difference formula \(K.E. = eV\), and then relate it to the velocity to find:
\(v = \frac{2eV}{m}\). This velocity is used later to balance the forces of the electric and magnetic fields.
Ion Acceleration
Ion acceleration occurs due to the potential difference applied across the ions, giving them kinetic energy. By accelerating the ions to a known velocity, we can predict their behavior in both electric and magnetic fields. In the exercise, the ions are initially accelerated via a \(10 \text{kV}\) potential difference, contributing directly to their kinetic energy:
\(K.E. = eV\). This acceleration ensures that the ions acquire sufficient velocity to allow analysis of their behavior in the magnetic field:
\(v = \frac{2eV}{m}\). Thus, we can then equate forces due to electric and magnetic fields to find the appropriate electric field strength that balances the magnetic deflection.

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Most popular questions from this chapter

Half Circle In Fig \(29-29\), a charged particle moves into a region of uniform magnetic field \(B\), goes through half a circle, and then exits that region. The particle is either a proton or an electron (you must decide which). It spends \(130 \mathrm{~ns}\) within the region. (a) What is the magnitude \(|\vec{B}| ?\) (b) If the particle is sent back through the magnetic field (along the same initial path) but with \(2.00\) times its previous kinetic energy, how much time does it spend within the field?

A Wire Carries a Current A wire \(1.80 \mathrm{~m}\) long carries a current of \(13.0 \mathrm{~A}\) and makes an angle of \(35.0^{\circ}\) with a uniform magnetic field of magnitude \(B=1.50 \mathrm{~T}\). Calculate the magnitude of the magnetic force on the wire.

Television Tube Each of the electrons in the beam of a television tube has a kinetic energy of \(12.0 \mathrm{keV}\). The tube is oriented so that the electrons move horizontally from geomagnetic south to geomagnetic north. The vertical component of Earth's magnetic field points down and has a magnitude of \(55.0 \mu \mathrm{T}\). (a) In what direction will the beam deflect? (b) What is the magnitude of the acceleration of a single electron due to the magnetic field? (c) How far will the beam deflect in moving \(20.0 \mathrm{~cm}\) through the television tube?

Positron A positron with kinetic energy \(2.0 \mathrm{keV}\) is projected into a uniform magnetic field \(\vec{B}\) of magnitude \(0.10 \mathrm{~T}\), with its velocity vector making an angle of \(89^{\circ}\) with \(\vec{B}\). Find (a) the period, (b) the pitch \(p\), and (c) the radius \(r\) of its helical path.

Current Loop A single-turn current loop, carrying a current of \(4.00 \mathrm{~A}\), is in the shape of a right triangle with sides \(50.0,120\), and \(130 \mathrm{~cm}\). The loop is in a uniform magnetic field of magnitude \(75.0 \mathrm{mT}\) whose direction is parallel to the current in the \(130 \mathrm{~cm}\) side of the loop. (a) Find the magnitude of the magnetic force on each of the three sides of the loop. (b) Show that the total magnetic force on the loop is zero.

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