A particle with charge −− 5.90 nCnC is moving in a uniform magnetic field B⃗ =−(B→=−( 1.28 TT )k^k^. The magnetic force on the particle is m

Question

A particle with charge −− 5.90 nCnC is moving in a uniform magnetic field B⃗ =−(B→=−( 1.28 TT )k^k^. The magnetic force on the particle is measured to be F⃗ =−(F→=−( 3.70×10−7 NN )i^+()i^+( 7.60×10−7 NN )j^)j^.

(a) Calculate all the components of the velocity of the particle that you can from this information.
(b) Are there components of the velocity that are not determined by the measurement of the force? Explain.
(c) Calculate the scalar product v⃗ ⋅F⃗ . What is the angle between v⃗ and F⃗

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Thanh Thu 5 years 2021-08-14T21:56:41+00:00 1 Answers 83 views 0

Answers ( )

  1. thienthi
    0
    2021-08-14T21:58:16+00:00
    Reply

    Answer:

    Explanation:

    Given that,

    Charge q=-5.90nC

    Magnetic field B= -1.28T k

    And the magnetic force

    F =−( 3.70×10−7N )i+( 7.60×10−7N )j

    Let the velocity be V(xi + yj + zk)

    Then, the force is given as

    Note i×i=j×j×k×k=0

    i×j=k. j×i=-k

    j×k=i. k×j=-i

    k×i=j. i×k=-j

    The force in a magnetic field is given as

    F= q(v×B)

    −( 3.70×10−7N )i+( 7.60×10−7N )j =

    q(xi + yj + zk) × -1.28k

    −( 3.70×10−7N )i+( 7.60×10−7N )j=

    q( -1.28x i×k – 1.28y j×k – 1.28z k×k)

    −( 3.70×10−7N )i+( 7.60×10−7N )j=

    q( 1.28xj – 1.28y i )

    −( 3.70×10−7N )i+( 7.60×10−7N )j=

    q( -1.28y i + 1.28x j)

    So comparing comparing coefficients

    let compare x axis component

    -( 3.70×10−7N )i=-1.28qy i

    −3.70×10−7N = -1.28qy

    y= -3.7×10^-7/-1.28q

    y= -3.7×10^-7/-1.28×-5.90×10^-9)

    y=-48.99m/s

    y≈-49m/s

    Let compare y-axisaxis

    7.6×10−7N j = 1.28qx j

    7.6×10−7N = 1.28qx

    x= 7.6×10^-7/1.28q

    x= 7.6×10^-7/1.28×-5.90×10^-9

    x=-100.64m/s

    a. Then, the velocity of the x component is x= -100.64m/s

    b. Also, the velocity component of the y axis is y=-49m/s

    c. We will compute

    V•F

    V=-100.64i -49j

    F=−( 3.70×10−7 N )i+( 7.60×10−7 N )j

    Note

    i.j=j.i=0. Also i.i = j.j =1

    V•F is

    (-100.64i-49j)•−(3.70×10−7N)i+(7.60×10−7 N )j =

    3.724×10^-5 – 3.724×10^-5=0

    V•F=0

    d. Angle between V and F

    V•F=|V||F|Cosx

    0=|V||F|Cosx

    Cosx=0

    x= arccos(0)

    x=90°

    Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

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