A proton is moving in a circular orbit of radius 11.8 cm in a uniform magnetic field of magnitude 0.232 T directed perpendicular to the velocity of the proton. Find the orbital speed of the proton. Answer in units of m/s.
I came up with .118/2.83x10^-7 and it was incorrect.
Of course it is incorrect. Do you think an electron can possibly travel that slow speed?
Let me see your work.
(2pie(1.67x10^-27)/ (1.60x10^-19)x .232= 2.83x10^-7
.118/2.83x10^-7 wrong answer. I have entered three wrong answers and can not pass this if I can not get the right answer. Please help!
To solve this problem, you need to use the concept of centripetal force and the magnetic force acting on a charged particle moving in a magnetic field.
The centripetal force, given by the equation Fc = mv²/r, is responsible for keeping the proton in a circular orbit. The magnetic force, given by the equation Fm = qvB, acts perpendicular to the velocity (v) of the proton.
In this case, the magnetic force and the centripetal force are equal:
qvB = mv²/r
Here, q is the charge of the proton (which is equal to the elementary charge e, approximately 1.6 x 10^-19 C), m is the mass of the proton (approximately 1.67 x 10^-27 kg), v is the orbital speed of the proton, B is the magnitude of the magnetic field (0.232 T), and r is the radius of the circular path (11.8 cm = 0.118 m).
To solve for v, we can rearrange the equation:
qvB = mv²/r
v² = qB / (mr)
v = √ (qB / (mr))
Substituting the given values:
v = √((1.6 x 10^-19 C)(0.232 T) / ((1.67 x 10^-27 kg)(0.118 m)))
Calculating this expression:
v ≈ 1.9 x 10^5 m/s
Therefore, the orbital speed of the proton is approximately 1.9 x 10^5 m/s.