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A particle with a charge of +3. 0x10-6C traveling at a velocity of 300,000m/s towards the top of the page through a magnetic field of 1. 4T to the left side of the page. What is the force on the particle? Include magnitude and direction

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Answer:

Step-by-step explanation:

The force on a charged particle moving in a magnetic field can be calculated using the formula:

F = qvBsinθ

where

F is the force on the particle

q is the charge of the particle

v is the velocity of the particle

B is the magnetic field strength

θ is the angle between the velocity and magnetic field vectors.

In this case, the charge on the particle is +3.0x10^-6C, its velocity is 300,000m/s, and the magnetic field is 1.4T. The angle between the velocity and magnetic field vectors is 90° because the particle is moving perpendicular to the magnetic field.

So, plugging in the values to the formula, we get:

F = (3.0x10^-6C)(300,000m/s)(1.4T)sin90°

F = (3.0x10^-6C)(300,000m/s)(1.4T)(1)

F = 1.26x10^-2 N

Therefore, the force on the particle is 1.26x10^-2 N to the left of the page, perpendicular to the velocity of the particle.

User Raphael Rafatpanah
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