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A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. The magnitude of the magnetic field is increased. Compared to the initial radius of the circular path, the radius of the new path is which of the following

1_ larger

2_ smaller

3_equal in size

Respuesta :

sebasacevedop

2_ smaller

The magnitude of the force on the particle is given by the magnetic component of the Lorentz force:

[tex]F_m=qvBsen\theta[/tex]

Where is the particle's charge, v its speed, B the magnetic field and [tex]\theta[/tex] is the angle between the particle motion and the magnetic field. According to the Newton's second law, we have:

[tex]F_m=F_c[/tex]

Where [tex]F_c[/tex] is the centripetal force, replacing the values of thee forces and solving for r:

[tex]qvBsen\frac{\pi}{2}=m\frac{v^2}{r}\\r=\frac{mv}{qB}[/tex]

The radius is inversely proportional to the magnetic field. Therefore, if the magnetic field is increased, the radius of the new path is smaller.

okpalawalter8

Answer:

  • The radius of the new path is Smaller(Option 2)

Explanation:

comparing magnetic force and centripetal force

[tex]qvB = \frac{mv^2}{r}[/tex]

radius of motion is given by = [tex]\frac{mv}{qB}[/tex]

if B is increased the radius will decrease

Therefore,

The radius of the new path is smaller, which is option 2

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brainly.com/question/15685257

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