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Chapter 5: Problem 86

In Chapter 19 we will see that a charged particle can undergo uniform circular motion when acted on by a magnetic force and no other forces. (a) For that to be true, what must be the angle between the magnetic force and the particle's velocity? (b) The magnitude of the magnetic force on a charged particle is proportional to the particle's speed, \(F=k v .\) Show that two identical charged particles moving in circles at different speeds in the same magnetic field must have the same period. (c) Show that the radius of the particle's circular path is proportional to the speed.

Short Answer

Expert verified
Answer: The radius of the circular path is proportional to the speed of the charged particle.

Step by step solution

01

Understand the fundamental properties of magnetic force and circular motion

Magnetic force acts perpendicularly to the velocity of the charged particle, so it doesn't cause any change in the particle's speed, only the direction. The force causing uniform circular motion acts perpendicular to the velocity of the particle. This fundamental property will be useful when examining the angle between magnetic force and the particle's velocity for uniform circular motion.
02

Find the angle between the magnetic force and the particle's velocity

Since the magnetic force is causing uniform circular motion, it must act perpendicularly to the velocity of the particle. Thus, the angle between the magnetic force and the particle's velocity is 90 degrees.
03

Analyze identical charged particles in the same magnetic field

The magnetic force on a charged particle is given by \(F=k v\). Let the charge particles be \(q_1\) and \(q_2\) with velocities \(v_1\) and \(v_2\). Equal magnetic forces \(F_1\) and \(F_2\) on the particles will result in uniform circular motion with radius \(r_1\) and \(r_2\), and angular frequencies \(\omega_1\) and \(\omega_2\).
04

Derive the period formula for the particles

For the magnetic force causing uniform circular motion, we have: \(F_1 = k v_1 = m_1 \omega_1^2 r_1\) and \(F_2 = k v_2 = m_2 \omega_2^2 r_2\) Since the charge particles are identical, they have the same mass: \(m_1 = m_2\) Now, we can derive the period formula for each particle: \(T_1 = \frac{2\pi}{\omega_1}\) and \(T_2 = \frac{2\pi}{\omega_2}\)
05

Show that the particles have the same period

From the force equations, we can write: \(\frac{v_1}{\omega_1^2 r_1} = \frac{v_2}{\omega_2^2 r_2}\) Now, substitute the period formulas: \(\frac{v_1}{(2\pi/T_1)^2 r_1} = \frac{v_2}{(2\pi/T_2)^2 r_2}\) Simplifying: \(\frac{v_1 T_1^2}{r_1} = \frac{v_2 T_2^2}{r_2}\) Now, we know the radius is proportional to velocity, so we have: \(r_1 = kv_1\) and \(r_2 = kv_2\) Substitute for \(r_1\) and \(r_2\): \(\frac{v_1 T_1^2}{kv_1} = \frac{v_2 T_2^2}{kv_2}\) Simplifying this will result in: \(T_1^2 = T_2^2\) So, taking the square root of both sides: \(T_1 = T_2\) Therefore, both identical charged particles moving in circles at different speeds have the same period.
06

Show the radius is proportional to speed

From the force equation: \(F = kv = m \omega^2 r\) We know the period and angular frequency are related: \(\omega = \frac{2\pi}{T}\) Substitute this into the previous equation: \(F = kv = m \left(\frac{2\pi}{T}\right)^2 r\) Now, solve for \(r\): \(r = \frac{kvT^2}{4\pi^2m}\) Since \(k\), \(T\), and \(m\) are constants, we can denote the constant of proportionality by \(C\): \(r = Cv\) Therefore, the radius of the particle's circular path is proportional to the speed.

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Most popular questions from this chapter

A car drives around a curve with radius \(410 \mathrm{m}\) at a speed of $32 \mathrm{m} / \mathrm{s} .$ The road is not banked. The mass of the car is \(1400 \mathrm{kg} .\) (a) What is the frictional force on the car? (b) Does the frictional force necessarily have magnitude $\mu_{\mathrm{s}} N ?$ Explain.
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