# Homework Help: Proton in Magnetic field - am I correct?

1. Jan 19, 2004

### Poobel

Proton in Magnetic field -- am I correct?

One of my Physics ISU questions is as follows:
- A proton moves in a circular path perpendicular to a 1.10 T magnetic field. The radius of the path is 4.5 cm. calculate energy of the proton in eV

Well i had hard time doing it but here we go: my solution

mV^2/r=QrB V=QrB/m

V= (1.6x10^-16) x o.o45 m x 1.10 T) / 1.67x10^-27 kg

V= 4.7x10^6 m/s

Ek= 1/2mV^2 = 1/2 x 1.67x10^-27 kg x (4.7x10^6 m/s)^2
= 1.8x10^-14 J

Ek= 1.87x10^-14J x eV/1.6x10^-19J = (1.87x10^-14J)/ (1.6x10^-19J)

eV=1.3x10^5 eV=1.3x10^5

Fellow physisists is my solution correct?

Thank you

2. Jan 19, 2004

### Warr

Hmm...I got close to 1.17*10^5 eV

Here's how I did it:

We want to know the kinetic energy of the proton, so:

$$E_{k}=\frac{1}{2}mv^2$$

We know the mass of a proton is 1.67*10^-27, so the variable we need is the velocity at which it is travelling.

As this proton is moving in a circle in a magnetic field,

$$F_m=F_c$$ (where $$F_m$$ is the magnetic force and $$F_c$$ is the centripetal force

representing these with equations, we have

$$qvB=\frac{mv^2}{r}$$

now, we isolate v, since that is what we are trying to get

$$v=\frac{qBr}{m}$$

substituting into the kinetic energy equation, we get

$$E_{k}=\frac{1}{2}m(\frac{qBr}{m})^2$$

which reduces to

$$E_{k}=\frac{(qBr)^2}{2m}$$

substituting in,

$$E_{k}=\frac{((1.6*10^-19)(1.10)(0.045))^2}{2(1.67*10^-27)}$$

which gets 1.878*10^-14 J

since 1 eV = 1.6*10^-19 J, the answer in eV is

(1.878*10^-14)/(1.6*10^-19) = 117,375 eV (1.17*10^5)

There are two things about yours that confuse me. First off, the magnitude of the charge on a proton is the elementary charge, which is 1.6*10^-19, whereas you have it as 1.6*10^-16...and yet you get the same energy in joules. Then at the end when you do your conversion from joules to electron volts, there seems to be a discrepency of some kind, because (1.87*10^-14)/(1.6*10^-19) does not equal 1.3*10^5. Also, I would have rounded to 1.88*10^5 if I were rounding. Apart from that the solution the same as mine.

Last edited: Jan 19, 2004
3. Jan 19, 2004

### Poobel

Ok I'll look more into it

thanks a bunch, help is always appreciated

BTW that -16... I just copied it wrong from my notes

Hmm also you must've read my first post wrong since

Ek= 1.87x10^-14J x eV/1.6x10^-19J = (1.87x10^-14J)/ (1.6x10^-19J)

eV=1.3x10^5 eV=1.3x10^5

Last edited: Jan 19, 2004