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A proton moves perpendicularly to a magnetic field that has a magnitude of 4.20 x 10-2 T. What is the speed of the particle if the magnitude of the magnetic force on it is 2.40 x 10-14 N?​

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khannadeem30664

Answer:

V=3.57 × 10^6 m/s

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The speed of the proton in the magnetic field is 3.57×10⁶ m/s.

To determine the speed of the proton, we need to know about the magnetic force experienced by a charged particle in a magnetic field.

What is the magnetic force on a charge particle in a magnetic field?

  • From lorentz law, the magnetic force on a charge particle is given as

               F=q×v×B

  where, q= charge of the particle

              v= speed of the particle

              B= magnetic field

What will be the speed of the charge particle, proton, if the magnetic field and force are given?

  • Magnetic force(F)= q×v×B

       v=F/(q×B)

  • Here, B= 4.20×10⁻²T

                  F=2.40×10⁻¹⁴N

                  q= 1.6×10⁻¹⁹C

       Then, v= 2.40×10⁻¹⁴/ ( 4.20×10⁻²× 1.6×10⁻¹⁹)

                   = 3.57×10⁶ m/s.

Thus, we can conclude that the speed of the proton in the magnetic field is 3.57×10⁶ m/s.

Learn more about the magnetic force here:

https://brainly.com/question/25932320

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