Notices
Results 1 to 4 of 4
Like Tree1Likes
  • 1 Post By AndrewC

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

  1. #1 !HELP! Electricity & Magnetism (Proton Path in a Magnetic Field) MasteringPhysics Assignment (PART E) 
    Junior Member
    Join Date
    Feb 2018
    Posts
    5
    A proton (q = 1.6010^−19 C , m = 1.6710^−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) = 1.1010^5 m/s , v (y-direction) =0, and v (z-direction) = 2.1010^5 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?

    ANSWER: in the +y direction



    PART C: In addition to the magnetic field there is a uniform electric field in the +x-direction, E⃗ =( 2.5010^4 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.

    --

    Anyone know how to go about this because I'm getting this wrong and I tried tbh

    Please help if you can! It's due soon enough.

    Thanks!
    Reply With Quote  
     

  2. #2  
    Senior Member
    Join Date
    Sep 2017
    Posts
    278
    Quote Originally Posted by QuantumDynamite99 View Post

    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) = 1.1010^5 m/s , v (y-direction) =0, and v (z-direction) = 2.1010^5 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.

    --

    Anyone know how to go about this because I'm getting this wrong and I tried tbh

    Please help if you can! It's due soon enough.

    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.
    Last edited by AndrewC; 02-28-2018 at 04:01 PM.
    QuantumDynamite99 likes this.
    Reply With Quote  
     

  3. #3  
    Junior Member
    Join Date
    Feb 2018
    Posts
    5
    Thanks! I already solved it thanks to you, just needed to replace u=0 with u=1.10^5 m/s like you suggested!
    Reply With Quote  
     

  4. #4  
    Senior Member
    Join Date
    Sep 2017
    Posts
    278
    Quote Originally Posted by QuantumDynamite99 View Post
    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!
    Reply With Quote  
     

Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •