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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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Most popular questions from this chapter

The Earth is showered with particles from space known as muons. They have a charge identical to that of an electron but are many times heavier \(\left(m=1.88 \cdot 10^{-28} \mathrm{~kg}\right)\) Suppose a strong magnetic field is established in a lab \((B=0.50 \mathrm{~T})\) and a muon enters this field with a velocity of \(3.0 \cdot 10^{6} \mathrm{~m} / \mathrm{s}\) at a right angle to the field. What will be the radius of the resulting orbit of the muon?

The work done by the magnetic field on a charged particle in motion in a cyclotron is zero. How, then, can a cyclotron be used as a particle accelerator, and what essential feature of the particle's motion makes it possible? A blue problem number indicates a worked-out solution is available in the Student Solutions Manual. One \(\bullet\) and two \(\bullet\) indicate increasing level of problem difficulty.

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) \(-v / 1836\)

The magnitude of the magnetic force on a particle with charge \(-2 e\) moving with speed \(v=1.0 \cdot 10^{5} \mathrm{~m} / \mathrm{s}\) is \(3.0 \cdot 10^{-18} \mathrm{~N}\). What is the magnitude of the magnetic field component perpendicular to the direction of motion of the particle?

A 30 -turn square coil with a mass of \(0.250 \mathrm{~kg}\) and a side length of \(0.200 \mathrm{~m}\) is hinged along a horizontal side and carries a 5.00 -A current. It is placed in a magnetic field pointing vertically downward and having a magnitude of \(0.00500 \mathrm{~T}\). Determine the angle that the plane of the coil makes with the vertical when the coil is in equilibrium. Use \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\).
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