Answer to Question #249992 in Physics for Quillz

Question #249992

Particles each with mass m and charge q accelerate through a potential difference of V = 100 volts and enter a region of uniform magnetic field B, where their paths are circular with radius R = 0.05 m as shown in Fig. 9.12. The magnitude of the field is B = 6.75 x 10βˆ’4 T. a. Indicate the direction of B on a drawing. b. Derive the following relationship between the radius R and q, m, V, and B: 𝑅 2 = π‘š π‘ž 2𝑉 𝐡2 c. Find the value of the charge-to-mass ratio (q/m) of these particles and identify them. Justify your answer! F


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Expert's answer
2021-10-12T12:33:32-0400

Equate the forces acting on the particles: Lorentz force equals the centripetal force, then express r.


qvB=mv2r, r=mqβ‹…vB.qvB=m\frac{v^2}r,\\\space\\ r=\frac mqΒ·\frac{v}{B}.

Express the velocity in terms of voltage and charge using energy conservation:


v=2qVm.v=\sqrt{\frac{2qV}{m}}.


Hence:


r=mqβ‹…2VB2, r2=mqβ‹…2VB2.r=\sqrt{\frac{m}{q}Β·\frac{2V}{B^2}},\\\space\\ r^2=\frac{m}{q}Β·\frac{2V}{B^2}.


The value of the charge-to-mass is


qm=2V(rB)2=175.583β‹…109 C/kg,\frac qm=\frac{2V}{(rB)^2}=175.583Β·10^9\text{ C/kg},

which corresponds to charge-to-mass ratio for electron. You can check it by diving the charge of electron by its mass.


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