Answer to Question #249992 in Physics for Quillz

Question

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 10−4 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 = 𝑚 𝑞 2𝑉 𝐵2 c. Find the value of the charge-to-mass ratio (q/m) of these particles and identify them. Justify your answer! F

Answer 1

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

$$qvB=m\frac{v^2}r,\\\space\\ r=\frac mq·\frac{v}{B}.$$

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

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

Hence:

$$r=\sqrt{\frac{m}{q}·\frac{2V}{B^2}},\\\space\\ r^2=\frac{m}{q}·\frac{2V}{B^2}.$$

The value of the charge-to-mass is

$$\frac qm=\frac{2V}{(rB)^2}=175.583·10^9\text{ C/kg},$$

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