/*! 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 61 An ion source is producing \({ }... [FREE SOLUTION] | 91Ó°ÊÓ

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An ion source is producing \({ }^{6} \mathrm{Li}\) ions, which have charge \(+e\) and mass \(9.99 \times 10^{-27} \mathrm{~kg}\). The ions are accelerated by a potential difference of \(25 \mathrm{kV}\) and pass horizontally into a region in which there is a uniform vertical magnetic field of magnitude \(B=1.2 \mathrm{~T}\). Calculate the strength of the electric field, to be set up over the same region, that will allow the \({ }^{6} \mathrm{Li}\) ions to pass through without any deflection.

Short Answer

Expert verified
The electric field strength is approximately 1.17 x 10^6 V/m.

Step by step solution

01

Calculate the speed of the ions

The ions are accelerated by a potential difference \( V = 25 \text{ kV} = 25 \times 10^3 \text{ V} \). The kinetic energy \( K \) acquired by each ion is given by \( eV \), where \( e \) is the elementary charge \( 1.6 \times 10^{-19} \text{ C} \). Thus, \( K = 1.6 \times 10^{-19} \times 25 \times 10^3 \). Use the kinetic energy formula \( K = \frac{1}{2}mv^2 \) to solve for \( v \),\[ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 25 \times 10^3}{9.99 \times 10^{-27}}} \]. Calculate \( v \).
02

Calculate the magnetic force on the ions

The magnetic force \( F_m \) experienced by an ion moving at speed \( v \) in a magnetic field \( B \) is given by \( F_m = evB \) with \( \theta = 90^\circ \) (since the velocity is perpendicular to the magnetic field). Substitute in the values, \( e = 1.6 \times 10^{-19} \text{ C} \), \( B = 1.2 \text{ T} \), and use \( v \) from Step 1 to find \( F_m \).
03

Determine the electric field strength

For the ions to pass without deflection, the electric force \( F_e = eE \) needs to balance the magnetic force \( F_m \). Set \( F_e = F_m \), so \( eE = evB \). Rearrange to solve for the electric field strength \( E \):\[ E = vB \]. Use the value of \( v \) from Step 1 and \( B = 1.2 \text{ T} \) to find \( E \).
04

Final calculations and conclusion

Substitute all known values to get \( E = v \times 1.2 \). With \( v \) calculated as approximately \( 975349 \, \text{m/s} \) from step 1, calculate \( E \): \( E = 975349 \times 1.2 \). The result is the required electric field strength in volts per meter.

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Key Concepts

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

Electric Field Strength
Electric field strength is a measure of how strong an electric field is at a given point. Its primary role is to exert force on charges within the field, influencing their paths.
In this exercise, you want to calculate the electric field needed to balance out the magnetic force acting on the ions. This way, the ions won't veer off course when moving through the magnetic field.

The balancing force requires you to set the electric force equal to the magnetic force. The formula you'll use is:
  • Electric Force (\( F_e \)) = Magnetic Force (\( F_m \))
  • \( eE = evB \)
Rearranging gives you the strength of the electric field \( E = vB \). Here, \( v \) is the ion's velocity, which is derived from ion acceleration by the potential difference. Finally, you'll calculate \( E \) based on the speed and the magnetic field strength.
Magnetic Force
Magnetic force is exerted on a charge moving through a magnetic field. It acts perpendicular to both the direction of the magnetic field and the velocity of the particle.
The formula for magnetic force is:\[ F_m = evB \] Here, \( e \) is the charge of the ion, \( v \) is the ion's velocity, and \( B \) is the magnetic field magnitude.

In this context, \( F_m \) acts horizontally; to ensure ions move straight, the magnetic force is countered by an equivalent electric force. The exercise helps you understand how balancing these two forces allows ions to pass through the region without deflection, aligning with the concept of cross product force in physics.
Ion Acceleration
Ion acceleration occurs when an ion experiences a potential difference. This acceleration increases the ion's velocity by converting electrical potential energy into kinetic energy.
The ions in this exercise are subjected to a potential difference of \( 25 \text{ kV} \). When accelerated across this voltage, the ions gain speed, calculated by the formula:
* Kinetic energy (\( K \)) = \( eV \)* Use this to find \( v \): \[ v = \sqrt{\frac{2eV}{m}} \] where \( m \) is the ion's mass.
By calculating \( v \), you determine how fast ions will travel, combining elements of mechanics and electromagnetism to describe their behavior in fields.
Kinetic Energy
Kinetic energy relates to the movement of an object, which is particularly relevant as ions accelerate through the potential difference. Here, you consider how the gained energy translates to increased speed.
For each ion, the kinetic energy is obtained through the expression:
\[ K = \frac{1}{2}mv^2 \]
Given earlier data, kinetic energy (\( K \)) is also related to the potential difference by \( K = eV \).

Understanding these equations lets students connect the foundational idea that energy conservation processes play a vital role in how ions move within magnetic and electric fields, especially in practical applications like accelerators and detectors.
Potential Difference
The potential difference, also known as voltage, is a key factor in accelerating ions; it denotes the work done per unit charge as it moves between two points in an electric field.
In this context, the chosen potential difference of \( 25 \text{ kV} \) provides the initial energy boost to the ions, causing them to start moving with a certain velocity.

These accelerated, charged particles are then subject to analysis by observing their movement in combined fields. Such an approach provides insights into controlling particle motion, crucial for diverse applications from industrial to healthcare sectors.

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Most popular questions from this chapter

Figure \(28-27\) shows a wire ring of radius \(a=1.8 \mathrm{~cm}\) that is perpendicular to the general direction of a radially symmetric, diverging magnetic fleld. The magnetic field at the ring is everywhere of the same magnitude \(B=3.4 \mathrm{mT}\), and its direction at the ring everywhere makes an angle \(\theta=15^{\circ}\) with a normal to the plane of the ring. The twisted lead wires have no effect on the problem. Find the magnitude of the force the field exerts on the ring if the ring carries a current \(i=4.6 \mathrm{~mA}\).

An electron is accelerated from rest by a potential difference of \(380 \mathrm{~V}\). It then enters a uniform magnetic field of magnitude \(200 \mathrm{mT}\) with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.

Figure 28-22 shows a metallic block, with its faces parallel to coordinate axes. The block is in a uniform magnetic field of magnitude \(0.020 \mathrm{~T}\). One edge length of the block is \(32 \mathrm{~cm}\); the block is not drawn to scale. The block is moved at \(3.5 \mathrm{~m} / \mathrm{s}\) parallel to each axis, in turn, and the resulting potential difference \(V\) that appears across the block is measured. With the motion parallel to the \(y\) axis, \(V=12 \mathrm{mV}\); with the motion parallel to the \(z\) axis, \(V=18 \mathrm{mV}\); with the motion parallel to the \(x\) axis, \(V=0\). What are the block lengths (a) \(d_{x}\), (b) \(d_{y}\), and (c) \(d_{z} ?\)

A long, rigid conductor, lying along an \(x\) axis, carries a current of \(7.0 \mathrm{~A}\) in the negative \(x\) direction. A magnetic field \(\vec{B}\) is present, given by \(\vec{B}=3.0 \mathrm{i}+8.0 x^{2} \hat{\mathrm{j}}\), with \(x\) in meters and \(\vec{B}\) in milliteslas. Find, in unit-vector notation, the force on the \(2.0 \mathrm{~m}\) segment of the conductor that lies between \(x=1.0 \mathrm{~m}\) and \(x=3.0 \mathrm{~m}\).

A magnetic dipole with a dipole moment of magnitude \(0.020 \mathrm{~J} / \mathrm{T}\) is released from rest in a uniform magnetic field of magnitude \(46 \mathrm{mT}\). The rotation of the dipole due to the magnetic force on it is unimpeded. When the dipole rotates through the orientation where its dipole moment is aligned with the magnetic field, its kinetic energy is \(0.80 \mathrm{~mJ}\). (a) What is the initial angle between the dipole moment and the magnetic field? (b) What is the angle when the dipole is next (momentarily) at rest?

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