A proton moving with a speed of 200 m/s enters perpendicular to the direction of the magnetic field and experiences a force of 2x10^-18N. What is the magnitude of the magnetic field?
any help is reallya appreciated
force=q B v
To find the magnitude of the magnetic field, you can use the equation:
F = qvb
where F is the force experienced by the particle, q is the charge of the particle, v is its velocity, and b is the magnetic field strength.
Given that the force experienced by the proton (q) is 2x10^-18 N and its velocity (v) is 200 m/s, we can substitute these values into the equation to solve for the magnetic field strength (b).
2x10^-18 N = (1.6x10^-19 C) * (200 m/s) * b
1.6x10^-19 C is the charge of a proton.
Now we can solve for b:
b = (2x10^-18 N) / (1.6x10^-19 C * 200 m/s)
b ≈ 6.25x10^-6 T
Therefore, the magnitude of the magnetic field is approximately 6.25x10^-6 Tesla.