A proton moving along the positive x-axis enters a uniform magnetic field which is directed along the

2 answers:

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Answer: The force acting on the proton of charge 1.5 x10-19 C moving with velocity 1.5 x10-19 C under the influence of a magnetic field of 0.7 T will be 3.15×10^-13 N.

Explanation: To find the answer we need to know more about the Lorentz magnetic force.

<h3>What is the Lorentz magnetic force acting on the proton?</h3>

Consider a proton of charge q moving with a velocity v in a magnetic field, then the Lorentz magnetic force exerted on the proton can be expressed as,

F= q (v× B)

$F= qvBsin\alpha$ where, $\alpha$ is the angle between v and B.

In the question, it is given that,

$B=0.7 T\\q=1.5*10^{-19}C\\v= 3*10^{6}m/s.\\\alpha =90 degree.\\$ because, from the question it is clear that the proton is moving along x axis and the magnetic field is along the y axis.

Thus, we can find the force acting on the proton as,

$F=qvBsin\alpha =1.5*10^{-19}C*3*10^6 m/s*0.7T*sin (90)\\F=3.15*10^{-13}N$

Thus, we can conclude that the Lorentz force acting on the proton will be 3.15×10^-13 N.

Learn more about the Lorentz magnetic force here:

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ivann1987 [24]2 years ago

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In response to a magnetic field of 0.7 T, a proton with a charge of 1.5 x 10-19 C travelling at a speed of 1.5 x 10-19 C will experience a force of 3.15 x 10^-13 N.

We need to learn more about the Lorentz magnetic force in order to locate the solution.

<h3>What does the proton experience as the Lorentz magnetic force?</h3>

If you imagine a proton with charge q travelling at speed v in a magnetic field, you can write down the Lorentz magnetic force acting on the proton as,

F= Q (v× B)

$F=QvBsin\alpha$

$\alpha$ the angle between v and B is, where.

It is stated in the query that, the magnetic field is along the y axis and the proton is travelling along the x axis. Thus, the angle will be 90 degrees.