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.
(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!
Bqv=mv^2/r
v=Bqr/m
v= .232*1.6E-19*.118/1.67E-27
Check that. Now compare it with whatever you did, I cant figure out what you did.
for information, if you posted this as your answer: .118/2.83x10^-7
I don't understand what you did, and why did you not simplify this. I get about 4.17E5 m/s from that fraction.
Check my work and see what I got from my calcs.
Calculating your v gave me 2.62285030e-48
so I then divide .118/ans and get 4.4989e46 would this be right?
To find the correct answer, we can use the formula for the centripetal force experienced by a charged particle moving in a magnetic field:
F = q * v * B
Where:
F is the centripetal force,
q is the charge of the particle (in this case, the charge of a proton, 1.60 x 10^-19 C),
v is the velocity of the particle (orbital speed in this case), and
B is the magnitude of the magnetic field (0.232 T).
The centripetal force is provided by the magnetic force acting on the proton:
F = m * a
Where m is the mass of the proton (1.67 x 10^-27 kg), and a is the centripetal acceleration.
Since the proton is moving in a circular orbit, the centripetal force can also be written as:
F = m * (v^2 / r)
Where r is the radius of the orbit (11.8 cm = 0.118 m).
Setting these two expressions for F equal to each other, we can solve for the orbital speed v:
q * v * B = m * (v^2 / r)
Rearranging the equation:
v = (m * r * B / q)^(1/2)
Substituting the given values:
v = (1.67 x 10^-27 kg * 0.118 m * 0.232 T / 1.60 x 10^-19 C)^(1/2)
Calculating this expression:
v = 2.92 x 10^6 m/s
Therefore, the orbital speed of the proton is approximately 2.92 x 10^6 m/s.
To find the orbital speed of the proton, we can make use of the equation for the centripetal force acting on a charged particle moving in a magnetic field:
F = (q * v * B), where F is the centripetal force, q is the charge of the particle, v is the velocity of the particle, and B is the magnitude of the magnetic field.
In this case, the centripetal force is provided by the magnetic force acting on the proton. Therefore, we have:
(q * v * B) = (m * v^2) / r, where m is the mass of the proton and r is the radius of the circular orbit.
Simplifying this equation, we can solve for the velocity:
v = (q * B * r) / m
Now, let's plug in the given values:
q = 1.60 x 10^-19 C (charge of the proton)
B = 0.232 T (magnitude of the magnetic field)
r = 11.8 cm = 0.118 m (radius of the orbit)
m = 1.67 x 10^-27 kg (mass of the proton)
v = (1.60 x 10^-19 C * 0.232 T * 0.118 m) / (1.67 x 10^-27 kg)
Calculating this expression, we get:
v ≈ 5.93 x 10^5 m/s
So, the orbital speed of the proton in this scenario is approximately 5.93 x 10^5 m/s.