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A particle of mass \(6.0 \times 10^{-8} \mathrm{kg}\) and charge \(+7.2 \mu \mathrm{C}\) is traveling due east. It enters perpendicularly a magnetic field whose magnitude is \(3.0 \mathrm{T}\). After entering the field, the particle completes one-half of a circle and exits the field traveling due west. How much time does the particle spend traveling in the magnetic field?

Short Answer

Expert verified
The particle spends approximately 0.00873 seconds in the magnetic field.

Step by step solution

01

Determine the Radius of Circular Path

When a charged particle enters a magnetic field perpendicularly, it travels in a circular path. The radius \( r \) of this path is given by the relation \( r = \frac{mv}{qB} \), where \( m \) is the mass, \( v \) is the velocity, \( q \) is the charge, and \( B \) is the magnetic field. To proceed, we need the velocity \( v \), but since it is not provided, we will assume it cancels out in the latter calculation for time.
02

Calculate Circumference of the Circle

The circumference \( C \) of a circle with radius \( r \) is given by \( C = 2 \pi r \). Since we are considering half of the circle, \( C/2 = \pi r \). We use this expression later on to relate it to velocity while calculating time.
03

Relate Velocities and Time

Knowing that time \( t \) to travel half the circle is given by \( t = \frac{\text{Distance}}{\text{Velocity}} \), replace \( \text{Distance} \) with \( \pi r \). Therefore, \( t = \frac{\pi r}{v} \). In previous steps, we'd usually cancel velocities if other relationships aren't given. Still, the initial velocity impacts this, hence suppose the calculations provide a time without explicit velocity input.
04

Input Known Values into Final Equation

To find time \( t \), input relevant quantities: \( r = \frac{mv}{qB} \) and the Time formula \( t = \frac{\pi r}{v} \). With known values: \( m = 6.0 \times 10^{-8} \mathrm{kg} \), \( q = 7.2 \times 10^{-6} \mathrm{C} \), \( B = 3.0 \mathrm{T} \). The velocity \( v \) will cancel out, so we solve: \( t = \frac{\pi \cdot \frac{6.0 \times 10^{-8}}{7.2 \times 10^{-6} \times 3.0}}{1} \). Therefore, \( t = \pi \cdot \frac{6.0 \times 10^{-8}}{21.6 \times 10^{-6}} \), simplifying it to \( t = \pi \times \frac{1}{360} \) s.
05

Calculate Time Numerically

To get numerical solutions substitute values \( t = \pi \times \frac{1}{360} \) seconds: \( \pi \approx 3.1416 \), \( t = 3.1416\times \frac{1}{360} = 0.00873 \) seconds.

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Key Concepts

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

Circular Motion in Magnetic Field
When a charged particle travels perpendicularly into a magnetic field, it moves in a circular path. This curious motion occurs because the magnetic field exerts a force that is always perpendicular to the velocity of the particle. This force is known as the magnetic Lorentz force. It continuously changes the direction of the particle, causing it to trace a circle.
This is analogous to how a stone tied to a string moves in a circle when you swing it around your head. The string tension keeps changing the stone’s path. Similarly, the magnetic force maintains the particle's circular motion.
  • The magnetic field does not do any work on the particle, so the speed of the particle remains constant.
  • The particle moves in a plane perpendicular to the magnetic field direction.
  • This consistent perpendicular interaction is why we observe circular motion.
Understanding these fundamental characteristics will help avoid confusion when studying similar problems.
Radius of Curvature
The radius of curvature is how large or small the circular path of the particle turns out to be. For a charged particle in a magnetic field, the radius depends on several factors: the particle’s mass ( m ), charge ( q ), velocity ( v ), and the strength of the magnetic field ( B ).
The formula to calculate the radius is:
\[ r = \frac{mv}{qB} \]
  • A larger radius means the particle makes wider turns.
  • If the particle is faster or heavier, it would have a larger radius.
  • A stronger magnetic field or higher charge leads to tighter, smaller circles.
This means that by adjusting these parameters, we can determine how tightly or loosely the particle circles through the magnetic field. This principle is key in devices like cyclotrons used for particle acceleration.
Charge and Magnetic Field Interaction
The interaction between a charged particle and a magnetic field is a fascinating showcase of electromagnetic principles. This interaction depends not only on the charge itself but also on the direction and magnitude of the velocity and the magnetic field.
Here’s what happens:
  • When the charged particle enters the magnetic field, it experiences a force that is perpendicular to both the field and its velocity.
  • This force is calculated by the cross product of velocity and the magnetic field, represented as: \( F = q(v \times B) \).
  • This force causes the particle to deviate from its path and engage in circular motion as discussed earlier.
Importantly, since the force is perpendicular to the motion, it changes the direction of the velocity, not its speed. This is why the interaction results in circular paths without speed alteration, aside from cases where other forces come into play.
Time of Motion in Magnetic Field
When examining how long a charged particle spends in a magnetic field, understanding the time it takes to complete a path is essential for grasping the full cycle of the motion.
To find the total time ( t ) the particle stays in the field, consider the formula:
\[ t = \frac{\pi r}{v} \]
Given that the radius ( r ) is determined from the velocity, charge, mass, and magnetic field strength using the relation \( r = \frac{mv}{qB} \), we can substitute \( r \) directly into the formula to figure out \( t \), effectively cancelling the velocity.
When simplified, this formula indicates that the time depends on the particle's physical properties and magnetic field conditions, irrespective of the initial velocity assumption.
  • This elegant result reveals deep symmetries in electromagnetic theory.
  • The time is particularly relevant for designing devices involving charged particle paths, like beam splitters and magnetic spectrometers.
Gaining confidence in using these relations will enhance your ability to predict and analyze charged particle behavior in magnetic applications.

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Most popular questions from this chapter

The magnetic field produced by the solenoid in a magnetic resonance imaging (MRI) system designed for measurements on whole human bodies has a field strength of \(7.0 \mathrm{T}\), and the current in the solenoid is \(2.0 \times 10^{2} \mathrm{A} .\) What is the number of turns per meter of length of the solenoid? Note that the solenoid used to produce the magnetic field in this type of system has a length that is not very long compared to its diameter. Because of this and other design considerations, your answer will be only an approximation.

The drawing shows a parallel plate capacitor that is moving with a speed of \(32 \mathrm{m} / \mathrm{s}\) through a \(3.6-\mathrm{T}\) magnetic field. The velocity \(\overrightarrow{\mathbf{v}}\) is perpendicular to the magnetic field. The electric field within the capacitor has a value of \(170 \mathrm{N} / \mathrm{C},\) and each plate has an area of \(7.5 \times 10^{-4} \mathrm{m}^{2} .\) What is the magnetic force (magnitude and direction) exerted on the positive plate of the capacitor?

Review Conceptual Example 2 as an aid in understanding this problem. A velocity selector has an electric field of magnitude \(2470 \mathrm{N} / \mathrm{C}\), directed vertically upward, and a horizontal magnetic field that is directed south. Charged particles, traveling east at a speed of \(6.50 \times 10^{3} \mathrm{m} / \mathrm{s},\) enter the velocity selector and are able to pass completely through without being deflected. When a different particle with an electric charge of \(+4.00 \times 10^{-12} \mathrm{C}\) enters the velocity selector traveling east, the net force (due to the electric and magnetic fields) acting on it is \(1.90 \times 10^{-9} \mathrm{N},\) pointing directly upward. What is the speed of this particle?

Particle 1 and particle 2 have masses of \(m_{1}=2.3 \times 10^{-8} \mathrm{kg}\) and \(m_{2}=5.9 \times 10^{-8} \mathrm{kg},\) but they carry the same charge \(q .\) The two particles accelerate from rest through the same electric potential difference \(V\) and enter the same magnetic field, which has a magnitude \(B\). The particles travel perpendicular to the magnetic field on circular paths. The radius of the circular path for particle 1 is \(r_{1}=12 \mathrm{cm} .\) What is the radius (in \(\mathrm{cm}\) ) of the circular path for particle \(2 ?\)

The two conducting rails in the drawing are tilted upward so they each make an angle of \(30.0^{\circ}\) with respect to the ground. The vertical magnetic field has a magnitude of \(0.050 \mathrm{T}\). The \(0.20-\mathrm{kg}\) aluminum rod (length \(=\) \(1.6 \mathrm{m}\) ) slides without friction down the rails at a constant velocity. How much current flows through the rod?

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