The Lorentz force causes charged particles to orbit around magnetic field lines. At what rate do protons orbit around a field line? Assume the protons have energy of 1 MeV and are in a magnetic field with strength B = 2.4×10-7 T. Find the orbital frequency in revolutions/second (Hz); express answer to 3 significant digits. The protons orbit the field at _____ Hz.
The Lorentz force causes charged particles to orbit around magnetic field lines. At what rate do protons orbit around a field line? Assume the protons have energy of 1 MeV and are in a magnetic field with strength B = 2.4×10-7 T. Find the orbital frequency in revolutions/second (Hz); express answer to 3 significant digits.
The protons orbit the field at _____ Hz.
Trending now
This is a popular solution!
Step by step
Solved in 4 steps with 4 images
How would the answer to the previous problem change if the particles were 1 MeV electrons, instead of protons?
by Bartleby ExpertAI answered the problem this way. Is this correct?
The Lorentz force that makes a charged particle move in a circular path in a magnetic field is given by
F=qvBF=qvB
where qq is the charge of the particle, vv is the speed of the particle, and BB is the magnetic field strength.
Since the particle is moving in a circular path, this force must equal the
Setting the two expressions for FF equal to each other, we get
qvB=mv2rqvB=rmv2
Since we want to find the frequency of the orbit, we need to find the speed vv of the proton. Given that the energy of the proton is 1 MeV, we can write
12mv2=1 MeV21mv2=1MeV
and solve for vv:
v=2×1 MeVmv=m2×1MeV
where mm is the mass of a proton, which is approximately 1.67×10−27 kg1.67×10−27kg.
Substituting this expression for vv into the expression for the Lorentz force, we can solve for the frequency of the orbit. The frequency is related to the speed and the radius of the path by
f=v2πrf=2πrv
Substituting our expressions for vv and rr, we find
f=qB2πmf=2πmqB
Substituting the given values for the charge of a proton (q=1.6×10−19 Cq=1.6×10−19C), the magnetic field strength (B=2.4×10−7 TB=2.4×10−7T), and the mass of a proton, we get
f≈(1.6×10−19 C)(2.4×10−7 T)2π(1.67×10−27 kg)≈(1.6)(2.4×10−7)2π(1.67×10−27)×106 eVf≈2π(1.67×10−27kg)(1.6×10−19C)(2.4×10−7T)≈2π(1.67×10−27)(1.6)(2.4×10−7)×106eV
f≈3.84×10−73.14×1.67×10−27×106 eVf≈3.14×1.67×10−273.84×10−7×106eV
f≈3.843.14×1.67×1014 Hzf≈3.14×1.673.84×1014Hz
f≈3.845.24×1014 Hzf≈5.243.84×1014Hz
f≈0.733×1014 Hzf≈0.733×1014Hz
f≈73.3×1012 Hzf≈73.3×1012Hz
So the protons orbit the field at 73.3×1012 Hz73.3×1012Hz, or 73.3 THz73.3THz.
by Bartleby Expert
