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Chapter 9: Problem 118

A charged particle of mass \(2 \mathrm{~kg}\) and charge \(2 \mathrm{C}\) move \(s\) with a velocity \(\vec{v}=8 \hat{i}+6 \hat{j} \mathrm{~ms}^{-1}\) in a magnetle field \(\vec{B}=2 \hat{k} \mathbf{T}\). Then a. the path of particle may be \(x^{2}+y^{2}=25\) b. the path of particle may be \(x^{2}+z^{2}=25\) c. the time period of particle will be \(3.14 \mathrm{~s}\) d none of these

Short Answer

Expert verified
The correct answer is both (a) and (c) are possible.

Step by step solution

01

Determine the Magnetic Force on the Particle

The magnetic force acting on the charged particle moving in a magnetic field is given by the Lorentz force equation \( \vec{F} = q(\vec{v} \times \vec{B}) \), where \( q \) is the charge of the particle, \( \vec{v} \) is the velocity, and \( \vec{B} \) is the magnetic field. Substituting the given values, \( \vec{v} = 8 \hat{i} + 6 \hat{j} \) and \( \vec{B} = 2 \hat{k} \), we find \( \vec{v} \times \vec{B} = (8 \hat{i} + 6 \hat{j}) \times 2 \hat{k} = (12 \hat{i} \times \hat{k} - 16 \hat{j} \times \hat{k}) = -12 \hat{j} + 16 \hat{i} \). Therefore, the force is \( \vec{F} = 2 (-12 \hat{j} + 16 \hat{i}) = -24 \hat{j} + 32 \hat{i} \).
02

Analyze the Path of the Particle

Since the force is perpendicular to the velocity and in the \(xy\)-plane (as it has components \(-24 \hat{j} + 32 \hat{i} \) and no \(\hat{k}\) component), the particle moves in the \(xy\)-plane. The path of the particle will be a circle in the \(xy\)-plane.
03

Determine the Radius of the Circular Path

The equation of motion for a charged particle in a magnetic field is centripetal, given by \( F = \frac{mv^2}{r}\), where \( v \) is the magnitude of the velocity. Using \( F = 40 \) from the vector magnitude \( (\sqrt{(-24)^2 + (32)^2} = 40) \): \( 40 = \frac{2 \times 10}{r} \). Solving for \( r \), we get \( r = 5 \). Thus \( x^2 + y^2 = 25 \) corresponds to this path.
04

Calculate the Time Period of Motion

The time period \( T \) of a charged particle in a magnetic field is given by \( T = \frac{2 \pi m}{qB} \). Plugging in the values \( m = 2 \), \( q = 2 \), and \( B = 2 \), we find \( T = \frac{2 \pi \times 2}{2 \times 2} = \pi \). This simplifies to approximately \( 3.14 \text{ s} \). Thus, option \( c \) is not correct as the calculation provided matches the statement but with corrected logic.

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Key Concepts

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

Lorentz Force
The Lorentz force is a fundamental concept when dealing with charged particles in magnetic fields. It describes the force experienced by a charged particle when it moves through an electric or magnetic field. In the context of this exercise, we're focusing on the magnetic part of the Lorentz force, given by the equation: \( \vec{F} = q (\vec{v} \times \vec{B}) \).
  • \( q \) is the charge of the particle.
  • \( \vec{v} \) is the velocity vector of the particle.
  • \( \vec{B} \) is the magnetic field vector.
The cross product \( (\vec{v} \times \vec{B}) \) means that the force is perpendicular to both the velocity and the magnetic field. In our exercise, the velocity \( \vec{v} = 8 \hat{i} + 6 \hat{j} \) and the magnetic field \( \vec{B} = 2 \hat{k} \) result in a force of \( -24 \hat{j} + 32 \hat{i} \), which lies entirely in the \( xy \)-plane. This perpendicular force is responsible for the circular motion of the particle within that plane.
Circular Motion in Magnetic Field
The movement of a charged particle in a magnetic field is often circular because the Lorentz force continuously acts perpendicular to its velocity, causing it to move in a circular path. Since this force acts continuously at right angles, it does not change the speed of the particle, only its direction, resulting in circular motion. When examining circular motion, we must look at two main concepts:
  • Centripetal Force: The requisite force for circular motion is provided entirely by the magnetic Lorentz force.
  • Radius of the Circle: Given by the formula \( r = \frac{mv}{qB} \), where \( m \) is the mass, \( v \) is the speed of the particle, \( q \) is the charge, and \( B \) is the magnetic field strength.
In the exercise, the magnitude of force \( F = 40 \) leads to the calculation of the radius \( r \) as 5 meters. Therefore, and as confirmed by solving the motion of the particle, the correct circular path equation is \( x^2 + y^2 = 25 \), where 25 is the square of the radius, confirming a circular path in the \( xy \)-plane.
Time Period of Charged Particle
The time period of a charged particle moving in a uniform magnetic field reflects how long it takes for the particle to complete one full circle. It is independent of both velocity and radius and is determined entirely by intrinsic properties of the particle and the magnetic field. The formula for the time period \( T \) is given by:\[ T = \frac{2 \pi m}{qB} \]
  • \( m \) represents the mass of the particle.
  • \( q \) is the charge.
  • \( B \) stands for the magnetic field strength.
This formula shows that the time period is directly proportional to the mass and inversely proportional to the charge and the magnetic field strength. It鈥檚 interesting to note that the time period does not depend on the velocity, making it the same regardless of how fast the particle moves within its circular path. In our specific problem, with given values \( m = 2 \), \( q = 2 \), and \( B = 2 \), the calculated time period comes out as \( T = \pi \) seconds, which is approximately 3.14 seconds. This calculation confirms the predicted time period based on the context of this charged particle's motion in the magnetic field.

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Most popular questions from this chapter

A current carrying loop lies on a smooth horizontal plane. Then, a it is possible to establish a uniform magnetic field in the region so that the loop starts rotating about its own axis b. it is possible to establish a uniform magnetic field in the region so that the loop will tip over about any of the point c. it is not possible that loop will tip over about any of the point whatever be the direction of established magnetic field (uniform) d both (a) and (b) are correct.

A charged particle moves with velocity \(\vec{v}=a \hat{i}+d f\) in a magnetic field \(\bar{B}=A \hat{i}+D \hat{j}\). The force acting on the particle has magnitude \(F .\) Then. a. \(F=0\), if \(a D=d A\). b. \(F=0\), if \(a D=-d A\). c. \(F=0\), if \(a A=-d D\). d. \(F \mu\left(a^{2}+b^{2}\right)^{1 / 2} \times\left(A^{2}+D^{2}\right)^{1 / 2}\)

A charged particle begins to move from the origin in a region which has a uniform magnetic field in the \(x\) -direction and a uniform electric field in the \(y\) -direction. Its speed is \(y\) when it reaches the point \((x, y, z)\). Then, \(v\) will depend a. only on \(x\) b. only on \(y\) c. on both \(x\) and \(y\), but not \(z\) d. on \(x, y\) and \(z\)

A particle of charge per unit mass \(\alpha\) is released from origin with a velocity \(v=\vec{v}_{0} \hat{i}\) in a uniform magnetic field \(\vec{B}=-B_{0} \hat{k}\). If the particle passes through \((0, y, 0)\), then \(y\) is equal to a- \(-\frac{2 v_{0}}{B_{0} \alpha}\) b \(\frac{v_{0}}{B_{0} \alpha}\) c. \(\frac{2 v_{0}}{B_{0} \alpha}\) \(\mathrm{d}=\frac{v_{0}}{B_{0} \alpha}\)

A current of \(1 /(4 \pi)\) ampere is flowing in a long straight conductor. The line integral of magnetic induction around \(\mathrm{a}\) closed path enclosing the current carrying conductor is a. \(10^{-7} \mathrm{~Wb} \mathrm{~m}^{-1}\) b \(4 \pi \times 10^{-7} \mathrm{~Wb} \mathrm{~m}^{-1}\) c. \(16 \pi^{2} \times 10^{-1} \mathrm{~Wb} \mathrm{~m}^{-1}\) d zero
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