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Group of particles in a magnetic field

  1. Jun 24, 2017 #1
    1. The problem statement, all variables and given/known data
    A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 km/s in the +x-direction experiences a force of 2.25 x ##10^{-16}##N in the +y-direction, and an electron moving at 4.75 km/s in the -z-direction experiences a force of 8.50 x ##10^{-16}##N in the +y-direction. a) What are the magnitude and direction of the magnetic field?

    2. Relevant equations
    Right hand rule
    F = |q|vB
    ##\vec F = qv##x##\vec B##

    3. The attempt at a solution
    info for proton:
    ##m_1 = 1.6727 * 10^{-27} kg##
    ##q_1 = +1.602 * 10^{-19} C##
    ##\vec v_1 = 1500 m/s## in the +x direction.
    ##\vec F_1 = 2.25 * 10^{-16} N## in the +y direction.

    info for the electron:
    ##m_2 = 9.110 * 10^{-31} kg##
    ##q_2 = -1.602*10^{-19} C##
    ##\vec v_2 = 4750 m/s## in the -z direction
    ##\vec F_2 = 8.50 * 10^{-16} ## in the +y direction

    ,i'm pretty sure I solved for B correctly incase you want to skip this part..
    ##\vec B = <b_1, b_2, b_3>##


    then i do cross product to solve for the B components.

    first equation, for proton:
    ##<0, 2.25 * 10^{-16}, 0> = q<1500, 0, 0>##x##<b_1, b_2, b_3>##
    second equation, for electron:
    ##<0,8.50*10^{-16}, 0> = -q<0, 0, -4750>##x##<b_1, b_2, b_3>##
    where ##q = +1.602*10^{-19}##

    first equation simplifies to this:
    ##q<1500, 0, 0>##x##<b_1, b_2, b_3>##
    ##= q<0, -1500b_3, 1500b_2>##
    ## = <0, -1500b_3q, 1500b_2q##

    second equation simplifies to this:
    ##-q<0, 0, -4750>##x##<b_1, b_2, b_3>##
    ##= -q<4750b_2, -4750b_1, 0>##
    ##= <-4750b_2q, 4750b_1q, 0>##

    so now i have 2 equations:
    ##<0, 2.25 * 10^{-16}, 0> = <0, -1500b_3q, 1500b_2q##
    ##<0,8.50*10^{-16}, 0> = <-4750b_2q, 4750b_1q, 0>##

    ##b_3 = \frac {2.25*10^{-16}} {-1500q} = -.936##
    ##b_2 = 0##
    ##b_1 = \frac {8.50*10^{-16}} { 4750q} = 1.117##

    ##B = \sqrt {(-.936)^2 + (1.117)^2} = 1.46T##

    now when i try to do the right hand rule.. i get confused
    what i know is velocity direction is the thumb, force direction is the palm, and magnetic field direction are the fingers.

    but there are 2 velocities and 2 forces... so i try to do it for both of them..

    for the proton my thumb points toward me, my palm faces to my right, and so my fingers point toward the floor. So the magnetic field is South.

    for the electron my thumb points toward the floor, my palm faces to my right, and so my fingers point away from me. So the magnetic field is in the -x direction.

    How can I find the direction?
     
  2. jcsd
  3. Jun 24, 2017 #2
    First off, the two sets of forces and velocities must not contradict each other, this is a good way to ensure your answer is correct. You are to check each of them separately. What you said about the right hand rule is correct, and so is your calculation so let's go over the two particles:

    (Important to note that the cross product is additive, so each of the two cross products can be broken down into a sum or cross products between the unit vectors. As you've calculated the B field has a positive x component and a negative z component, this already tells you the direction: somewhere between the x and z axises in that xz plane, the angle itl be from the x axis will be arctan(b3/b1).)


    To check this field actually exerts the forces given we can use the right hand rule:

    proton: Velocity is in the x direction and so is your thumb. now we look at the 2 components of the B-field:
    b1: the x component is parallel to the velocity and therefore has no effect, cross product between two parellel vectors always gives 0.
    b3: thumb with the x axis and fingers at negative z gives positive y direction for your palm, as given in the question

    electron: velocity is in the negative z direction and so is your thumb. examine both components of the vector:
    b1: thumb to negative x and fingers (b1=B component in the x direction that you found) to the x direction leaves your palm facing the negative y direction. recall the electron charge is negative so we flip the direction (multiply by -1). we get the positive y direction again for the force, as given in the question.
    b3: this component is parallel to the velocity and therefore has no effect.

    make sure you're using a right handed cartesian coordinate system and that you are in fact using the right hand ! (i made that mistake in an exam once..)

    Another way to see it: the cross product of two vectors ⃗v = ⟨v1, v2, v3⟩ and w⃗ = ⟨w1, w2, w3⟩ in space is




    defined as the vector:
    ⃗v×w⃗ =⟨v2w3 −v3w2,v3w1 −v1w3,v1w2 −v2w1⟩.
     
  4. Jun 25, 2017 #3
    Ok.. I followed your right hand rule steps to find the direction, thank you
     
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