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Chapter 27: Problem 8

An electron moves in a circular trajectory with radius \(r_{\mathrm{i}}\) in a constant magnetic field. What is the final radius of the trajectory when the magnetic field is doubled? a) \(\frac{r_{i}}{4}\) b) \(\frac{r_{i}}{2}\) c) \(r_{i}\) d) \(2 r_{i}\) e) \(4 r_{\mathrm{i}}\)

Short Answer

Expert verified
Question: When the magnetic field in which an electron is moving in a circular trajectory is doubled, the final radius of the trajectory is: a) Twice the initial radius b) Half the initial radius c) The same as the initial radius d) Four times the initial radius Answer: b) Half the initial radius

Step by step solution

01

Setting the magnetic force and centripetal force equal to each other

We have two equations: \(F = qvB\) (magnetic force) and \(F = \frac{mv^2}{r}\) (centripetal force). Since the magnetic force is providing the centripetal force for the electron, we can set these two equations equal to each other: $$ \Rightarrow qvB = \frac{mv^2}{r} $$
02

Solving for the radius in terms of the magnetic field

Now, we can solve for the radius \(r\) in terms of the magnetic field \(B\). Rearrange the equation as follows: $$ \Rightarrow r = \frac{mv}{qB} $$
03

Doubling the magnetic field

We are given that the magnetic field is doubled. Let \(r_i\) be the initial radius and \(r_f\) be the final radius. We can write two equations: $$ \begin{cases}r_i = \frac{mv}{qB_i}\\ r_f = \frac{mv}{q(2B_i)} \end{cases} $$
04

Finding the relationship between initial and final radii

Now, to find the relationship between \(r_i\) and \(r_f\), we can divide the second equation by the first equation: $$ \frac{r_f}{r_i} = \frac{\frac{mv}{q(2B_i)}}{\frac{mv}{qB_i}} = \frac{B_i}{2B_i} = \frac{1}{2} $$
05

Solving for the final radius

We have found the relationship between the initial and final radii, so we can now solve for the final radius \(r_f\) in terms of the initial radius \(r_i\): $$ \Rightarrow r_f = \frac{1}{2} r_i $$ The final radius of the trajectory when the magnetic field is doubled is \(\frac{r_i}{2}\). Therefore, the correct answer is option (b).

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Key Concepts

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

Centripetal Force
Centripetal force is a crucial concept when dealing with circular motion. It is the force that keeps an object moving along a curved path, directed towards the center of curvature. When an electron moves in a circular path within a magnetic field, the centripetal force is provided by the magnetic force. This is why their equations can be set equal to one another: \( F = qvB = \frac{mv^2}{r} \).
  • \(F\) is the force acting towards the center.
  • \(m\) is the mass of the electron.
  • \(v\) is the speed of the electron.
  • \(r\) is the radius of the trajectory.
  • \(q\) is the charge of the electron.
  • \(B\) is the magnetic field strength.
By understanding centripetal force, you can see why increasing the magnetic field affects the radius. Doubling the magnetic field increases the force acting on the electron, which pulls it more strongly toward the center, reducing the radius of the path.
Electron Trajectory
An electron trajectory refers to the path that an electron takes through space. When an electron is in a magnetic field, its trajectory is typically a circular path. The direction and radius of this path depend on several factors, including the velocity of the electron and the strength of the magnetic field.If the magnetic field increases, as in our initial exercise, electrons will tend to have a smaller radius of curvature as they are pulled more tightly into their circular paths. The key equation here is: \[ r = \frac{mv}{qB} \]This showcases that the trajectory's radius \( r \) decreases when the magnetic field \( B \) increases, holding everything else constant. When the magnetic field is doubled, the radius becomes half, demonstrating the inverse relationship between them. Thus, understanding this principle is paramount when predicting how an electron will move in varying magnetic fields.
Magnetic Force
Magnetic force affects charged particles such as electrons moving through magnetic fields. It acts perpendicular to both the direction of the magnetic field and the velocity of the particle. This is central to the creation of a circular path for the electron.The formula for magnetic force is: \[ F = qvB \] - \(F\) represents the magnetic force.- \(q\) is the charge of the particle.- \(v\) is the velocity of the particle.- \(B\) is the strength of the magnetic field.The magnetic force is key to maintaining the electron in motion along its curved path by providing the necessary centripetal force. When the magnetic field is strengthened, the force increases, causing the trajectory radius to shrink as seen in the exercise. Understanding the effects of magnetic force enhances comprehension of electron behavior in different field strengths.

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Most popular questions from this chapter

A proton moving at speed \(v=1.00 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) enters a region in space where a magnetic field given by \(\vec{B}=\) \((-0.500 \mathrm{~T}) \hat{z}\) exists. The velocity vector of the proton is at an angle \(\theta=60.0^{\circ}\) with respect to the positive \(z\) -axis. a) Analyze the motion of the proton and describe its trajectory (in qualitative terms only). b) Calculate the radius, \(r\), of the trajectory projected onto a plane perpendicular to the magnetic field (in the \(x y\) -plane). c) Calculate the period, \(T,\) and frequency, \(f\), of the motion in that plane. d) Calculate the pitch of the motion (the distance traveled by the proton in the direction of the magnetic field in 1 period).

A semicircular loop of wire of radius \(R\) is in the \(x y\) -plane, centered about the origin. The wire carries a current, \(i\), counterclockwise around the semicircle, from \(x=-R\) to \(x=+R\) on the \(x\) -axis. A magnetic field, \(\vec{B}\), is pointing out of the plane, in the positive \(z\) -direction. Calculate the net force on the semicircular loop.

A conducting rod of length \(L\) slides freely down an inclined plane, as shown in the figure. The plane is inclined at an angle \(\theta\) from the horizontal. A uniform magnetic field of strength \(B\) acts in the positive \(y\) -direction. Determine the magnitude and the direction of the current that would have to be passed through the rod to hold it in position on the inclined plane.

A high electron mobility transistor (HEMT) controls large currents by applying a small voltage to a thin sheet of electrons. The density and mobility of the electrons in the sheet are critical for the operation of the HEMT. HEMTs consisting of AlGaN/GaN/Si are being studied because they promise better performance at higher powers, temperatures, and frequencies than conventional silicon HEMTs can achieve. In one study, the Hall effect was used to measure the density of electrons in one of these new HEMTs. When a current of \(10.0 \mu\) A flows through the length of the electron sheet, which is \(1.00 \mathrm{~mm}\) long, \(0.300 \mathrm{~mm}\) wide, and \(10.0 \mathrm{nm}\) thick, a magnetic field of \(1.00 \mathrm{~T}\) perpendicular to the sheet produces a voltage of \(0.680 \mathrm{mV}\) across the width of the sheet. What is the density of electrons in the sheet?

An electron (with charge \(-e\) and mass \(m_{\mathrm{e}}\) ) moving in the positive \(x\) -direction enters a velocity selector. The velocity selector consists of crossed electric and magnetic fields: \(\vec{E}\) is directed in the positive \(y\) -direction, and \(\vec{B}\) is directed in the positive \(z\) -direction. For a velocity \(v\) (in the positive \(x\) -direction \(),\) the net force on the electron is zero, and the electron moves straight through the velocity selector. With what velocity will a proton (with charge \(+e\) and mass \(m_{\mathrm{p}}=1836 \mathrm{~m}_{\mathrm{e}}\) ) move straight through the velocity selector? a) \(v\) b) \(-v\) c) \(v / 1836\) d) \(-v / 1836\)
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