# Charge/Mass Ratio of Electron & Induction??

Hi, there. I'm learning Magnetic Field and Charge/Mass ratio and I'm so lost right now. Is there someone who can help me check whether I'm doing right or not? I'll so thank you.

1) The electron moves in a circle of radius r = 0.051cm. The magnetic field

B = 0.64T. Determine the speed of the electron. The mass and charge of the electron

can be found in the textbook. How many times is the speed of the electron smaller

than speed of light c = 3.0 × 10^8m/s?

In my solution, I used r = mv/eB. I isolated v so v = reB/m = (0.00051)(0.64)(1.6*10^-19) / (9.11*10^-31) = 5.73 *10^7 m/s. Then v/c = 5.73 *10^7 / 3.0 × 10^8 = 0.19 Thus, the speed of the electron is 0.19 times smaller then light.

2) The magnetic field due to Earth is approximately equal to 0.5×10^-4T. What current

is required in Helmholtz coils used in the lab to produce magnetic field at least ten

times larger than the magnetic field due to Earth.

B = 10.27*10^-4 * I(current) But then the magnetic field is ten times larger than Earth, 0.5*10^-4 *10 = 10.27*10^-4 * I

So I = (0.5 * 10^-3) / (10.27*10^-4) = 0.487A.

Could you tell me if there is anything wrong with my calculation?

### 2 Answers

- NCSLv 72 months agoFavorite Answer
F = q * v x B

and the force F = m*a = mv²/r

and since v and B are at right angles, v x B has magnitude v*B:

so

mv²/r = q*v*B

v = q*B*r / m

I get

v = 5.73e7 m/s

agree completely so far

n = c / v = 5.2 times smaller than light.

I think that this awful wording is throwing you off. (Or it's throwing me off.)

2) There is nothing wrong with your calculation here. But I don't know how you got to

B = 10.27*10^-4 * I

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- VamanLv 72 months ago
I missed. I do not know the radius of the coil.

B=I/r. B=10.27*10^(-4). We need to know r, the radius of the coil.

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