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Chapter 9: Problem 28

A particle of mass \(m\) and charge \(+q\) is shot with velocity \(\mathbf{v}\) into a region of uniform magnetic field \(\mathbf{B}\). Suppose that \(\mathbf{v}\) points to the top of the page and \(\mathbf{B}\) points out of the page. Ignoring gravity, the particle (A) travels in a straight line. (B) moves clockwise in a circle with radius \(r=q B / m v\). (C) moves counterclockwise in a circle with radius \(r=q B / m v\). (D) moves clockwise in a circle with radius \(r=m v / q B\). (E) moves counterclockwise in a circle with radius \(r=m v / q B\).

Short Answer

Expert verified
The motion of a positively charged particle in a uniform magnetic field when its velocity vector is pointing upwards and the magnetic field vector is pointing out of the page can be described as follows: The particle moves in a clockwise circular path with a radius given by \(r = \frac{mv}{qB}\).

Step by step solution

01

Determine the force acting on the particle

According to the Lorentz force formula, the force acting on a charged particle in a magnetic field is given by \(\mathbf{F} = q(\mathbf{v} \times \mathbf{B})\). Since the particle's velocity points to the top of the page and the magnetic field points out of the page, we can use the right-hand rule to find the direction of the force acting on the particle.
02

Apply the right-hand rule

Place the thumb of your right hand in the direction of the velocity vector (upwards) while keeping your fingers in the direction of the magnetic field (out of the page). The curl of your fingers shows the direction of the force. In this case, it is to the right.
03

Determine the motion of the particle

Since a force is being applied to the particle, it will not travel in a straight line. As the force is always perpendicular to the velocity vector, the particle will move in a circular path. As the charge is positive, the force direction we found using the right-hand rule shows that the particle moves clockwise in a circle.
04

Calculate the radius of the circular path

In a circular path, the centripetal force acting on the particle is given by \(F_{centripetal} = \frac{mv^2}{r}\). The magnetic force acting on the particle is the centripetal force, so we can set them equal: \(q(v\times B) = \frac{mv^2}{r}\). We want to find the radius \(r\) of the circular path, so we can solve this equation for \(r\): \(r = \frac{mv}{qB}\). Thus, the correct answer is: (E) moves counterclockwise in a circle with radius \(r= \dfrac{mv}{qB}\).

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