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Chapter 21: Problem 27

A positively charged particle of mass \(7.2 \times 10^{-8} \mathrm{kg}\) is traveling due east with a speed of \(85 \mathrm{m} / \mathrm{s}\) and enters a \(0.31-\mathrm{T}\) uniform magnetic field. The particle moves through one- quarter of a circle in a time of \(2.2 \times 10^{-3} \mathrm{s},\) at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.

Short Answer

Expert verified
(a) The magnetic force is approximately 3.89 mN. (b) The charge's magnitude is approximately 148 碌C.

Step by step solution

01

Understand the Motion

The particle moves in a circular path due to the magnetic field, completing one-quarter of a circle in 2.2 milliseconds. Given the uniform motion, we will find the radius of this circle.
02

Calculate the Radius of Motion

To find the radius of the circular path, we need the relationship between time, speed, and radius. The time for one complete circle would be four times 2.2 ms, so the radius \(r\) can be determined by: \( T = \frac{2\pi r}{v} \). Solving for \(r\) gives \[ r = \frac{85 \times 2.2 \times 10^{-3} \times 4}{2\pi} = 0.12 \text{ m} \].
03

Apply Lorentz Force Formula

The magnetic force on a moving charge in uniform magnetic field is given by \( F = qvB \). Circular motion implies \( F = \frac{mv^2}{r} \). Equating gives \( qvB = \frac{mv^2}{r} \), so solve for \(F\): \[ F = \frac{mv^2}{r} \].
04

Calculate the Magnetic Force

Substitute the known values into the force equation: \( m = 7.2 \times 10^{-8} \text{ kg}, v = 85 \text{ m/s}, r = 0.12 \text{ m}\). \[ F = \frac{(7.2 \times 10^{-8} \cdot 85^2)}{0.12} \approx 3.89 \times 10^{-3} \text{ N} \].
05

Determine the Charge Magnitude

Using the relationship \( q = \frac{F}{vB} \), substitute \( F = 3.89 \times 10^{-3} \text{ N}, v = 85 \text{ m/s}, B = 0.31 \text{ T} \). Solve for \( q \): \[ q = \frac{3.89 \times 10^{-3}}{85 \times 0.31} \approx 1.48 \times 10^{-4} \text{ C} \].

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Key Concepts

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

Circular Motion in Magnetic Fields
When a charged particle enters a magnetic field perpendicularly, it experiences a force that causes it to move in a circular path. The magnetic field doesn鈥檛 change the speed of the particle, only its direction, thus creating circular motion. This is because the magnetic force is always perpendicular to the velocity of the particle, which is a necessary condition for circular motion.
  • The path of motion remains circular as long as the magnetic force is perpendicular to velocity.
  • The radius of this circular path is determined by the balance of this magnetic force and the centripetal force needed for circular motion.
  • The speed of the particle doesn鈥檛 change even though its direction does.
For instance, if a particle completes one-quarter of a circle in a given time, you can calculate the radius of the full path using the relationship between time, speed, and radius. The calculation here requires the known speed and time, which allows us to find how large the circle would be if it were completed.
Lorentz Force
The Lorentz Force is the principle that describes the magnetic force exerted on a moving charged particle in a magnetic field. The magnetic force \( F \) is calculated by the formula:\[ F = qvB \]where:- \( q \) is the charge of the particle, - \( v \) is its velocity, - \( B \) is the magnetic field strength.Since force is directed perpendicular to the motion in a uniform magnetic field, this is what causes circular motion. In our exercise, by recognizing that the magnetic and centripetal forces are equal, we equate them:\[ qvB = \frac{mv^2}{r} \]From this formula, we can find the magnetic force \( F \) exerted on the particle by rearranging it in terms of the given mass \( m \), speed \( v \), and radius \( r \).This force calculation is vital to understanding the dynamics of charged particles moving through magnetic fields in applications such as cyclotrons and other particle accelerators.
Charge Calculation
Once the magnetic force on the particle is calculated, determining the charge \( q \) of a particle becomes straightforward using rearranged Lorentz force equations. Given the magnetic force formula \( F = qvB \), solving for the charge gives:\[ q = \frac{F}{vB} \]Where:
  • \( F \) is the magnitude of the magnetic force obtained previously,
  • \( v \) is the velocity of the particle, and
  • \( B \) is the magnetic field strength.
By substituting known values into the formula, one can find the exact value of the charge. This method of calculating charge is practical in many physics problems involving magnetic fields, especially when we cannot measure the charge directly. Accurate calculation of the charge provides insight into the particle鈥檚 behavior and interactions with the magnetic field, playing a crucial role in scientific research and industrial applications such as mass spectrometry and electron beam dynamics.

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Most popular questions from this chapter

At a certain location, the horizontal component of the earth's magnetic field is \(2.5 \times 10^{-5} \mathrm{T},\) due north. A proton moves eastward with just the right speed for the magnetic force on it to balance its weight. Find the speed of the proton.

A \(45-\mathrm{m}\) length of wire is stretched horizontally between two vertical posts. The wire carries a current of \(75 \mathrm{A}\) and experiences a magnetic force of 0.15 N. Find the magnitude of the earth's magnetic field at the location of the wire, assuming the field makes an angle of \(60.0^{\circ}\) with respect to the wire.

Suppose that a uniform magnetic field is everywhere perpendicular to this page. The field points directly upward toward you. A circular path is drawn on the page. Use Amp猫re's law to show that there can be no net current passing through the circular surface.

A long solenoid has 1400 turns per meter of length, and it carries a current of 3.5 A. A small circular coil of wire is placed inside the solenoid with the normal to the coil oriented at an angle of \(90.0^{\circ}\) with respect to the axis of the solenoid. The coil consists of 50 turns, has an area of \(1.2 \times 10^{-3} \mathrm{m}^{2}\) and carries a current of 0.50 A. Find the torque exerted on the coil.

A charged particle with a charge-to-mass ratio of \(|q| / m=5.7 \times\). \(10^{8} \mathrm{C} / \mathrm{kg}\) travels on a circular path that is perpendicular to a magnetic field whose magnitude is 0.72 T. How much time does it take for the particle to complete one revolution?
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