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.

 

[asteriatic]

I would approach this problem by first identifying the key variables and equations that can be used to solve for the distance between adjacent turns of the helix. From the given information, we know the speed and direction of the proton, the strength and direction of the magnetic field, and the mass and charge of the proton.

The key equations that can be used to solve for the distance are the equations for centripetal force and the equation for the period of a circular motion. We can set these two equations equal to each other and solve for the radius, which will then allow us to find the distance between adjacent turns of the helix.

F = qvBsinθ = mv^2/r

r = mv/ qBsinθ

Next, we can use the equation for the period of a circular motion to find the time it takes for the proton to complete one full circle.

T = 2πr/v

With the radius and period, we can then use the equation xf = xi + vt to find the distance the proton travels in the x-direction during one full circle. This distance will be equal to the distance between adjacent turns of the helix.

Therefore, the distance between adjacent turns of the helix can be calculated as:

Distance between adjacent turns = xf - xi = 2πr - 0 = 2πr = 2π(mv/ qBsinθ)

Plugging in the given values, we get:

Distance between adjacent turns = 2π((1.67*10^-27 * 4300)/(1.602*10^-19 * 25*10^-6 * sin76°))

Distance between adjacent turns = 0.106 meters

Therefore, the distance between adjacent turns of the helix formed by the proton's path is approximately 0.106 meters.