Find distance proton moves between adjacent turns of helix

[jburt]

Homework Statement

A proton moves with a speed of 4300 m/s in a direction 76.0° above positive x axis. It enters a region where a magnetic filed of 25x10^-6 Tesla exists in the positive x direction. Find the radius of the helix formed by the protons path and the distance between adjacent turns of helix.

To find the distance, I'm not sure. I know I need to find distance around and time it takes for a complete circle.

Homework Equations

Force of mag. field = qvsin∅B
F=qvsin∅B=ma
F= qvsin∅B = mv^2/r

r = mvsin∅/ qB

v = 2∏r/t

t = 2∏r/v


xf=xi + vt
vyfinal = vyinitial + at

The Attempt at a Solution

r = (1.67*10^-27 * 4300sin76°) / (1.602*10^-19 * 25*10^-6)
r = 1.74m


t = 2∏r/v
t = .002542s

so, vyfinal = vyinitial + at

xf = 0 + 4300m/s(.002542m)

xf = 10.9, which is NOT the answer.

 

[gneill]

Homework Statement

A proton moves with a speed of 4300 m/s in a direction 76.0° above positive x axis. It enters a region where a magnetic filed of 25x10^-6 Tesla exists in the positive x direction. Find the radius of the helix formed by the protons path and the distance between adjacent turns of helix.

To find the distance, I'm not sure. I know I need to find distance around and time it takes for a complete circle.

Homework Equations

Force of mag. field = qvsin∅B
F=qvsin∅B=ma
F= qvsin∅B = mv^2/r

r = mvsin∅/ qB

v = 2∏r/t

t = 2∏r/v


xf=xi + vt
vyfinal = vyinitial + at

The Attempt at a Solution

r = (1.67*10^-27 * 4300sin76°) / (1.602*10^-19 * 25*10^-6)
r = 1.74m

Okay, you've used the velocity of the particle that is perpendicular to the field. Good. That should give you the correct radius.

t = 2∏r/v
t = .002542s

Something fishy here. Did you use the speed of the particle rather than its velocity component in the (moving) plane of the loop? Hint: The speed of the particle in the loop should be the same as that which was used in the calculation of the loop radius.

 

[jburt]

so I have Δy=vsin∅t + at^2

Δy= 2∏r

but vsin∅final = vsin∅initial + at

F = ma

qvsin∅B = ma

a = qvsin∅B/m

a = (25*10^-6T)(4300sin76m/s)(1.602*10^-19C)/ (1.67*10^-27kg) = 1.0006*10^7
t = vsin∅final/a = 4172.27/1.0006*10^7 = .000417s

xfinal = vinitial*t = 4300m/s*.000417s =1.793m but it's wrong according to my professors answers. 8-(

 

[gneill]

so I have Δy=vsin∅t + at^2

Δy= 2∏r

but vsin∅final = vsin∅initial + at

Can you add some explanation to the above formulas? What acceleration are you dealing with? As far as I can tell the only acceleration you need to worry about is that involved with the circular motion -- centripetal acceleration.

F = ma

qvsin∅B = ma

a = qvsin∅B/m

a = (25*10^-6T)(4300sin76m/s)(1.602*10^-19C)/ (1.67*10^-27kg) = 1.0006*10^7
t = vsin∅final/a = 4172.27/1.0006*10^7 = .000417s

I think you're missing a factor of $2\pi$ in the above calculation.

xfinal = vinitial*t = 4300m/s*.000417s =1.793m but it's wrong according to my professors answers. 8-(

The proton does not have the full speed of 4300m/s in the x-direction.