A proton has a speed of 3 x 10^6 m/s

[Dx]

Hello,
I need help deriving a formula to solve for this please.

A proton has a speed of 3 x 10^6 m/s in a direction perpendicular to a uniform magnetic field and the proton moves in a circle radius .2 m. what is the magnitude of the magnetic field?

If someone can show me the all formulas to use i can manipulate from here.
Thanks!
Dx

 

[HallsofIvy]

A particle with mass m, charge q, moving at speed v in a constant magnetic field of magnitude B at right angles to the velocity vector of the particle, will move in a circle with radius mv/(qB)

 

[M98Ranger]

To solve for the magnitude of the magnetic field, we can use the formula for the centripetal force on a charged particle in a magnetic field:

F = qvB

Where F is the centripetal force, q is the charge of the particle (in this case, the charge of a proton is 1.602 x 10^-19 C), v is the speed of the particle, and B is the magnitude of the magnetic field.

We also know that the centripetal force is equal to the product of the mass of the particle (in this case, the mass of a proton is 1.67 x 10^-27 kg), the speed squared (v^2), and the radius of the circle (r):

F = mv^2 / r

We can equate these two formulas and solve for B:

mv^2 / r = qvB
B = mv / qr

Plugging in the given values, we get:

B = (1.67 x 10^-27 kg)(3 x 10^6 m/s) / (1.602 x 10^-19 C)(.2 m)
B = 2.5 x 10^-2 T

Therefore, the magnitude of the magnetic field is 2.5 x 10^-2 Tesla. Keep in mind that this is the magnitude of the field, not the direction. The direction of the field can be determined using the right hand rule. I hope this helps!