A charged particle accelerated to a velocity v enters the chamber of a mass spectrometer. The particle's velocity is perpendicular to the direction of the uniform magnetic field B in the chamber. After the particle enters the magnetic field, its path is a:

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Answer:

Circle

Explanation:

When a charged particle is in motion in a region with magnetic field, the particle experiences a force whose magnitude is given by

[tex]F=qvB sin \theta[/tex]

where

q is the charge

v is the velocity of the particle

B is the strength of the magnetic field

[tex]\theta[/tex] is the angle between the directions of v and B

In this problem, the velocity of the particle is perpendicular to the magnetic field, so

[tex]\theta=90^{\circ}[/tex]

and the formula reduces to

[tex]F=qvB[/tex]

Also, the direction of this force is perpendicular to the direction of motion of the particle. This means that as the charge moves in the region of the magnetic field, the force acting on it acts as a centripetal force: therefore, the particle will start moving by unifom circular motion, with constant speed (because the magnetic force does no work on the particle, since it is perpendicular to the direction of motion).

So, the path of the particle will be a circle.

The path of the charged particle enters the uniform magnetic field is a circular or circle path.

What is magnetic field?

The magnetic field is the field in the space and around the magnet in which the magnetic field can be filled.

The magnetic field experienced by a charged particle can be given as,

[tex]B=\dfrac{F}{qv\sin\theta}[/tex]

Here, (q) is the charge of the particle, (v) is the speed of the particle, and (F) is the magnitude of the magnetic force.

A charged particle accelerated to a velocity v enters the chamber of a mass spectrometer. The particle's velocity is perpendicular to the direction of the uniform magnetic field B in the chamber.

For the perpendicular direction, the value of angle is equal to 90 degrees. Therefore, the above formula can be written as,

[tex]B=\dfrac{F}{qv\sin(90)}\\B=\dfrac{F}{qv}[/tex]

Here, the magnetic force acting on the charged particle is perpendicular to the magnetic field as well as the velocity of the particle. Thus, the path of the particle should be circular.

Hence, the path of the charged particle enters the uniform magnetic field is a circular or circle path.

Learn more about magnetic field here;

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