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Chapter 15: Problem 54

A proton of mass \(1.67 \times 10^{-27} \mathrm{~kg}\) and charge \(1.6 \times\) \(10^{-19} \mathrm{C}\) is projected with a speed of \(2 \times 10^{6} \mathrm{~m} / \mathrm{s}\) at an angle of \(60^{\circ}\) to the \(x\)-axis. If a uniform magnetic field of \(0.104 \mathrm{~T}\) is applied along \(y\)-axis, the path of proton is (A) a circle of radius \(=0.2 \mathrm{~m}\) and time period \(\pi \times 10^{-7} \mathrm{~s}\). (B) a circle of radius \(=0.1 \mathrm{~m}\) and time period \(2 \pi \times 10^{-7} \mathrm{~s}\). (C) a helix of radius \(=0.1 \mathrm{~m}\) and time period \(2 \pi \times 10^{-7} \mathrm{~s}\). (D) a helix of radius \(=0.2 \mathrm{~m}\) and time period \(4 \pi \times 10^{-7} \mathrm{~s}\).

Short Answer

Expert verified
The path of the proton is (C) a helix of radius \(=0.1 \mathrm{~m}\) and time period \(2 \pi \times 10^{-7} \mathrm{~s}\).

Step by step solution

01

Calculate the magnetic force exerted on the proton

To calculate the magnetic force(F), we can use the formula: \(F = qvBsin\theta \) where q is the charge of the particle, v is its velocity, B is the magnetic field strength, and \(\theta\) is the angle between the velocity and the magnetic field. Given, charge (q) = \(1.6\times10^{-19}\mathrm{C}\), velocity (v) = \(2\times10^6\mathrm{m/s}\), magnetic field (B) = \(0.104\mathrm{T}\), and angle (theta) = \(60^{\circ}\). Let's calculate the magnetic force. \(F = (1.6\times10^{-19})(2\times10^6)(0.104)sin(60^{\circ})\)
02

Find the radius of the circle or helix

Now we need to find the radius of the circular or helical path. We can use the formula: \(r = \frac{mv_{\perp}}{qB}\) where m is the mass of the particle, \(v_{\perp}\) is the component of velocity perpendicular to the magnetic field, q is the charge of the particle, and B is the magnetic field strength. Given, mass (m) = \(1.67\times10^{-27}\mathrm{kg}\), velocity (v) = \(2\times10^{6}\mathrm{m/s}\), angle \(\theta = 60^{\circ}\), magnetic field (B) = \(0.104 \mathrm{T}\), and charge (q) = \(1.6\times10^{-19}\mathrm{C}\). We can calculate \(v_{\perp}\) as \(v_{\perp} = v\sin\theta = (2\times10^{6})\sin(60^{\circ})\) Now, let's find the radius (r). \(r =\frac{(1.67\times10^{-27})(2\times10^{6})\sin(60^{\circ})}{(1.6\times10^{-19})(0.104)}\)
03

Find the time period of the path

Next, we need to find the time period(T) of the circular or helical motion. We can use the formula: \(T = \frac{2\pi m}{qB}\) where m is the mass of the particle, q is the charge of the particle, and B is the magnetic field strength. Given, mass (m) = \(1.67\times10^{-27}\mathrm{kg}\), magnetic field (B) = \(0.104\mathrm{T}\), and charge (q) = \(1.6\times10^{-19}\mathrm{C}\). Let's calculate the time period (T). \(T = \frac{2\pi (1.67\times10^{-27})}{(1.6\times10^{-19})(0.104)}\) Now compare the calculated values of radius and time period with the given options. The option that matches the calculations is the correct answer for the path of the proton.

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Key Concepts

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

Helical Motion
When a charged particle like a proton enters a magnetic field at an angle, it doesn't simply travel in a straight line or pure circle. Instead, it moves in what is called a helical motion. This is a combination of linear and circular motions.
The particle's velocity can be split into two components:
  • The component that is parallel to the magnetic field, which causes the particle to move linearly.
  • The component that is perpendicular to the magnetic field, which results in circular motion.
Both components work simultaneously, creating a helical path around the magnetic field lines.
The path is like that of a spring or a slinky stretched out along the direction of the magnetic field. This movement causes the charged particle to spiral around the field lines.
Because the component responsible for the circular motion does not depend on time, the radius of the helical path remains constant as long as the magnetic field remains unchanged.
Radius of Path
The radius of the path in the case of a helical or circular motion of a charged particle under a magnetic field is an important parameter. It is determined by using the equation: \[ r = \frac{mv_{\perp}}{qB} \] where:
  • \(m\) is the mass of the particle,
  • \(v_{\perp}\) is the component of the particle's velocity perpendicular to the magnetic field,
  • \(q\) is the charge of the particle,
  • \(B\) is the magnetic field strength.
In simpler terms, it measures how broadly the circle component of the helical path spreads. The larger the radius, the less tightly the particle spirals.
For instance, if a proton enters the magnetic field and makes an angle of \(60^{\circ}\) with the axis, only part of its velocity is perpendicular to the field. We need to compute this perpendicular component, \(v \sin \theta\), to calculate the radius.
This calculation ensures us understand how the charged particle delineates its motion not only in the magnetic field but also how factors like mass, charge, and field strength interplay in determining this radius.
Time Period of Motion
The time period of the motion of a charged particle in a uniform magnetic field is the time it takes for a complete path or loop of its motion. The formula to determine the time period \(T\) is given by: \[ T = \frac{2\pi m}{qB} \] where:
  • \(m\) is the mass of the particle,
  • \(q\) is the charge of the particle,
  • \(B\) is the magnetic field strength.
This formula shows that the time period is independent of the velocity of the particle.
Instead, it depends on the mass and charge of the particle and the strength of the magnetic field. As a result, each charged particle in the same magnetic field can have a very predictable time period while moving along their helical paths.
When considering scenarios in homework problems, if you compute the time period correctly, you can easily compare it against given options to determine the correct description of the particle's path. This is particularly useful when trying to distinguish between motion paths such as helical versus circular under different conditions.

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Most popular questions from this chapter

A current \(i\) ampere flows in a circular arc of wire whose radius is \(R\), which subtend an angle \(3 \pi / 2\) radian at its centre. The magnetic induction \(B\) at the centre is (A) \(\frac{\mu_{0} i}{R}\) (B) \(\frac{\mu_{0} i}{2 R}\) (C) \(\frac{2 \mu_{0} i}{R}\) (D) \(\frac{3 \mu_{0} i}{8 R}\)

A conducting \(\operatorname{rod} A B\) of length \(l=1 \mathrm{~m}\) is moving at a velocity \(v=4 \mathrm{~m} / \mathrm{s}\) making an angle \(30^{\circ}\) with its length. \(C\) is the middle point of the rod. A uniform magnetic field \(B=2 \mathrm{~T}\) exists in a direction perpendicular to the plane of motion, then (A) \(v_{A}-v_{B}=8 \mathrm{v}\) (B) \(v_{A}-v_{B}=4 \mathrm{v}\) (C) \(v_{B}-v_{C}=2 \mathrm{v}\) (D) \(v_{B}-v_{C}=-2 \mathrm{v}\)

Two very long, straight, parallel wires carry currents \(I\) and \(-I\), respectively. The distance between the wires is \(d\). At a certain instant of time, a point charge \(q\) is at a point equidistant from the two wires, in the plane of the wires. Its instantaneous velocity \(\vec{v}\) is perpendicular to this plane. The magnitude of the force due to magnetic field acting on the charge at this instant is (A) \(\frac{\mu_{0} I q v}{2 \pi d}\) (B) \(\frac{\mu_{0} I d q}{\pi d}\) (C) \(\frac{2 \mu_{0} I q v}{\pi d}\) (D) zero

Two short bar magnets of magnetic moments \(M\) each are arranged at the opposite corners of a square of side \(d\) such that their centres coincide with the corners and their axes are parallel. If the like poles are in the same direction, the magnetic induction at any of the other corners of the square is (A) \(\frac{\mu_{0}}{4 \pi} \frac{M}{d^{3}}\) (B) \(\frac{\mu_{0}}{4 \pi} \frac{2 M}{d^{3}}\) (C) \(\frac{\mu_{0}}{4 \pi} \frac{M}{2 d^{3}}\) (D) \(\frac{\mu_{0}}{4 \pi} \frac{M^{2}}{2 d^{3}}\)

Three identical bar magnets, each of magnetic moment \(M\), are placed in the form of an equilateral triangle with north pole of one touching the south pole of the other as shown. The net magnetic moment of the system is (A) Zero (B) \(3 M\) (C) \(\frac{3 M}{2}\) (D) \(M \sqrt{3}\)
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