What is the Magnitude of the Magnetic Field for a Proton Beam?

[EvanQ]

Homework Statement

A beam of protons traveling at 1.20 km/s enters a uniform magnetic field, traveling perpendicular to the field. The beam exits the magnetic field, leaving the field in a direction perpendicular to its original direction. The beam travels a distance of 1.30 cm while in the field.

What is the magnitude of the magnetic field?

Homework Equations

F = qv ×B
dF = Idl ×B

The Attempt at a Solution

F = qv x B
B = F/qv
= F / (1.602 × 10−19) 1.2

stuck as to what equation applies, and where i can find the value of F from.

 

[Midy1420]

the second equation deals with forces on current carrying wires...so you would deal with the first. F is a force...what else in physics is F (Force) equal to...also think of the motion of the proton...it follows a _________ path use that to equate force.

 

[EvanQ]

ok so:

F = qv x B
B = F/qv
= F / (1.602 × 10^−19) x 1.2 (charge of a proton x velocity of proton)

F = ma?
= 1.67x10^-27 x a (mass of a proton x acceleration)

however, as:
m(v/t) / q(v)
= m(1/t) / q
= (1.67x10^-27 x (1/t)) / 1.602 × 10^−19 (where t = distance / velocity??)

 

[EvanQ]

i got 1.129x10^-13 and failed my first attempt.

 

[Midy1420]

how can you relate acceleration with what you are given in the problem we are given "v" and we are also given "r". the acceleration is purely centripetal so the velocity does not change

 

[Midy1420]

the only other thing is it doesn't say it made a radius of 1.3 so you would have to find it using arc length formula

I hope this is helping you

 

[EvanQ]

yeh i was about to ask how you knew the radius. ok i'll work on those and get back to you in a few mintues.

**digs deep into the memory bank for equations**

 

[Astronuc]

See if this reference helps jog the memory -

http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/maspec.html#c2

In a quarter turn, l = pi/2 * r

 

[EvanQ]

would i be correct in saying that using the arc length formula:

pi/2 radians is equivalent to 1.3cm
θ = s / r
r = s / θ
= 1.3 / pi/2
r = 0.8276 cm

t = 0.013m / 1200m/s
= 1.083x10^-5 s

then w = θ / t
= (pi/2) / 1.083x10^-5
= 145041.2121

w = w0 + at
145041.2121 = 0 + a x 1.083x10^-5
a = 1.339x10^10 ??

therefore:

B = F/qv
= ma / qv
= (1.67x10^-27 x 1.339x10^10) / ((1.602×10^−19) x 1200)
= 0.1163 T

or using w instead of v, 9.6237x10^-4put both in and got it wrong again sigh, where am i going wrong?

 

[EvanQ]

r = mv/qb

0.008276 = (1.67x10^-27 x 1200) / (1.602×10^−19 x b)

(1.602×10^−19 x b) = (1.67x10^-27 x 1200) / 0.008276)

(1.602×10^−19 x b) = 2.42146x10^-22

b = 1.5115x10^-3

 

[EvanQ]

that was correct :)

thanks heaps for your help

 

[rl.bhat]

Here the force is centripetal force.
SO qvB= mv^2/r
B = mv/qr
substitute the values. See how much you get.