Electromagnetism-Radius of Curvature-Question

[RoyalFlush100]

Homework Statement

http://postimg.org/image/gdfsdglpt/

v=2.2E5 m/s
l=0.16 m
I=5.3 A
N=820
mp=1.67E-27 kg
B=3.4E-2 T

Looking for radius of curvature in the solenoid with the above variables.

Homework Equations

BIl=F

The Attempt at a Solution

BIl=mv^2/r
3.4E-2*5.3*0.16=1.67E-27*(2.2E5)^2/r
r=2.8E-15 m

The correct answer however, is 6.8E-2 m, so what did I do wrong?

 

[haruspex]

horizontal movement on the slope

BIl=mv^2/r
3.4E-2*5.3*0.16=1.67E-27*(2.2E5)^2/r

Think about what I and L refer to in BIL=F. You already used the length of the solenoid and the current in it to find the field. Does it make sense to be using those again to find the force?

 

[RoyalFlush100]

Think about what I and L refer to in BIL=F. You already used the length of the solenoid and the current in it to find the field. Does it make sense to be using those again to find the force?

Fm=Fc
Bqv=mv^2/r
r=mv/Bq
r=(1.67E-27*2.2E5)/(3.4E-2*1.6*10^-19)
r=6.8E-2 m

Thanks. Why didn't the other formula for force work though?

 

[haruspex]

Fm=Fc
Bqv=mv^2/r
r=mv/Bq
r=(1.67E-27*2.2E5)/(3.4E-2*1.6*10^-19)
r=6.8E-2 m

Thanks. Why didn't the other formula for force work though?

What you calculated previously seems to be the solenoid exerting a force on another like itself. This is the danger with learning formulae but not exactly what the variables in the formula represent. It's not enough that I is a current, it matters how that current relates to the magnetic field, etc.