A proton moving in a circular path perpendicular to a constant magnetic field takes 2.6 µs (2.6 × 10−6 s) to complete one revolution. The proton has a charge of 1.60218 × 10−19 C and a mass of 1.67262 × 10−27 kg. Determine the magnitude of the field. Answer in units of T.

The magnitude of the field is 0.025 Tesla.  

Explanation:

The magnetic field force over a charge particle is defined as:

 (1)

Where q is the charge, v is the velocity, B is the magnetic field and is the angle between the displacement of the particle and the magnetic field.

For this particular case, is equal to , since the proton is moving in a circular path perpendicular to a constant magnetic field.

Therefore, from equation 1 it is gotten:

 (2)

Then, replacing Newton's second law in equation 2 it is gotten:

However, a is the centripetal acceleration since the proton describes a circular motion:

(3)

Notice that the magnetic field, B, can be isolated from equation 3

  (4)

But v can be replaced with the orbital speed:

 (5)

Where T is the period.

Therefore, equation 5 can be replaced in equation 4

 (6)

Finally, equation 6 can be used with the values given:

But and    

Hence, the magnitude of the field is 0.025 Tesla.                    

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