# Thread: !HELP! Electricity & Magnetism (Proton Path in a Magnetic Field) MasteringPhysics Assignment (PART E)

1. A proton (q = 1.60×10^−19 C , m = 1.67×10^−27 kg )moves in a uniform magnetic field B⃗ =( 0.450 T )i^.
At t = 0, the proton has a velocity components v (x-direction) = 1.10×10^5 m/s , v (y-direction) =0, and v (z-direction) = 2.10×10^5 m/s .

PART A: What is the magnitude of the magnetic force acting on the proton?

ANSWER: Magnetic Force = 1.51 x10^-14

PART B: What is the direction of the magnetic force acting on the proton?

PART C: In addition to the magnetic field there is a uniform electric field in the +x-direction, E⃗ =( 2.50×10^4 V/m )i^. Will the proton have a component of acceleration in the direction of the electric field?

ANSWER: Acceleration in the x-direction = 2x40x10^12 m/s^2

PART D: Describe the path of the proton. Does the electric field affect the radius of the helix? Explain.

MY ANSWER: The magnetic force (in the +y direction) will cause the proton to make a helical motion perpendicular to the y-z plane, centripetal in nature. The magnetic field is in the x-direction contributing to the radius of the helix. No, the electric field does not affect the radius of the helix, as all the electric force is acting in the x-direction. The radius only depends on the magnetic field and charge of the particle which derives the formula, F = q * v * B.

PART E: At t=T/2, where T is the period of the circular motion of the proton, what is the x-component of the displacement of the proton from its position at t = 0?

MY ATTEMPT:

m: 1.67x10^-27 kg
q: 1.60x10^-19 C
a (x-direction): 2.40x10^12 m/s^2
B: 0.450 T
v (x-direction): 1x10^5 m/s

The formula I would've used was R= mv/qB
so that I would derive w=Bq/m
then knowing T=2pi/w
T=2pi*m/Bq

but T=t/2 from question of PART E,

so divided by 2 to get
t=pi*m/Bq

so I used this formula, plugged in the values and got 7.286749627x10^-8 seconds

Then plugged it in to the Newton equation, s=ut+1/2at^2 where u=0

so s=1/2at^2

but getting a from PART C to be 2.40x10^12 m/s^2, I plugged that in and got

s=1/2(2.40x10^12)(7.286749627x10^-8)^2

s= 6.371606415x10^-3 metres

which is d in the x-direction, yet I put it in to my online MasteringPhysics question, but got it wrong.

The answer is not 6.37x10^-3 metres.

--

Thanks!

2. Originally Posted by QuantumDynamite99

MY ANSWER: The magnetic force (in the +y direction) will cause the proton to make a helical motion perpendicular to the y-z plane, centripetal in nature. The magnetic field is in the x-direction contributing to the radius of the helix. No, the electric field does not affect the radius of the helix, as all the electric force is acting in the x-direction. The radius only depends on the magnetic field and charge of the particle which derives the formula, F = q * v * B.
You will need to get the equation of the trajectory (the "helix") in order to solve the next part (part E).

x(t)=....
y(t)=.....
z(t)=.....

For that, you will need to form and solve the equations of motion, a set of differential equations:

with the initial conditions at t = 0, v (x-direction) = 1.10×10^5 m/s , v (y-direction) =0, and v (z-direction) = 2.10×10^5 m/s

PART E: At t=T/2, where T is the period of the circular motion of the proton, what is the x-component of the displacement of the proton from its position at t = 0?

MY ATTEMPT:

m: 1.67x10^-27 kg
q: 1.60x10^-19 C
a (x-direction): 2.40x10^12 m/s^2
B: 0.450 T
v (x-direction): 1x10^5 m/s

The formula I would've used was R= mv/qB
so that I would derive w=Bq/m
then knowing T=2pi/w
T=2pi*m/Bq

but T=t/2 from question of PART E,

so divided by 2 to get
t=pi*m/Bq

so I used this formula, plugged in the values and got 7.286749627x10^-8 seconds

Then plugged it in to the Newton equation, s=ut+1/2at^2 where u=0
u is not 0, it is 1.1*10^5, see above.

so s=1/2at^2

but getting a from PART C to be 2.40x10^12 m/s^2, I plugged that in and got

s=1/2(2.40x10^12)(7.286749627x10^-8)^2

s= 6.371606415x10^-3 metres

which is d in the x-direction, yet I put it in to my online MasteringPhysics question, but got it wrong.

The answer is not 6.37x10^-3 metres.

--

Thanks!
You will need to calculate:

x(T/2)=....

If you solve the differential equation, you will find out that the answer is quite simple. Hint: x depends only on the electric field, E.

3. Thanks! I already solved it thanks to you, just needed to replace u=0 with u=1.10^5 m/s like you suggested!

4. Originally Posted by QuantumDynamite99
Thanks! I already solved it thanks to you, just needed to replace u=0 with u=1.10^5 m/s like you suggested!
Excellent!

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