Answer to Question #159801 in Electric Circuits for zain ul abdeen

Question #159801

 Singly charged uranium-238 ions are accelerated through a potential difference of 2.00 kV and enter a uniform magnetic field of 1.20 T directed perpendicular to their velocities. (a) Determine the radius of their circular path. (b) Repeat for uranium-235 ions. What If? How does the ratio of these path radii depend on the accelerating voltage? On the magnitude of the magnetic field? 


1
Expert's answer
2021-02-04T02:34:15-0500

a) Using the Newton’s Second Law of Motion we can write:


"qvB=\\dfrac{mv^2}{r},""r=\\dfrac{mv}{qB}."

We can find the velocity of the ions from the law of conservation of energy:


"qV=\\dfrac{1}{2}mv^2,""v=\\sqrt{\\dfrac{2qV}{m}}."

Substituting "v" into the formula for the radius of the uranium-238/235 ions orbit, we get:


"r=\\dfrac{\\sqrt{2Vm}}{\\sqrt{q}B}."


Then, we can find the radius of uranium-238 ions circular path:


"r_{238}=\\dfrac{\\sqrt{2\\cdot2\\cdot10^3\\ V\\cdot238.05\\cdot1.66\\cdot10^{-27}\\ kg}}{\\sqrt{1.602\\cdot10^{-19}\\ C}\\cdot1.20\\ T}=0.08278\\ m."

b) Let's repeat the calculations for uranium-235 ions:


"r_{235}=\\dfrac{\\sqrt{2\\cdot2\\cdot10^3\\ V\\cdot235.04\\cdot1.66\\cdot10^{-27}\\ kg}}{\\sqrt{1.602\\cdot10^{-19}\\ C}\\cdot1.20\\ T}=0.08225\\ m."

c) Let's find the ratio of the path of uranium-235/238 radii:


"\\dfrac{r_1}{r_2}=\\dfrac{\\dfrac{\\sqrt{2Vm_1}}{\\sqrt{q}B}}{\\dfrac{\\sqrt{2Vm_2}}{\\sqrt{q}B}}=\\sqrt{\\dfrac{m_1}{m_2}}."

As we can see from the formula, the ratio of these path radii doesn't depend on the accelerating voltage. Also, it doesn't depend on the magnitude of the magnetic field.


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