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Magneto-static force and Newton's 3rd Law

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  1. Aug 9, 2017 #1
    1. The problem statement, all variables and given/known data
    Show with an example whether Magnetic Force follows Newton's 3rd Law.

    2. Relevant equations


    3. The attempt at a solution
    Let's consider the following system at time t.
    upload_2017-8-10_8-41-45.png
    ##\vec B ## due to moving q1 at q2 i.e. ## \vec B_1 ## is along ## \hat y ## direction.
    So, the force on the q2 is ## \vec F_{on~q2} = q_2 v B_1 \hat z ##

    Taking q2 = q1, it's clear that the strength of magnetic field due to q2 at q1 i.e. B_2 is very small compared to B_1.
    So, it's clear that magneto - static force doesn't follow Newton's 3rd Law.


    This results because both magnetic field due to a moving charge and the magnetic force acting on a moving charge are perpendicular to its velocity.
    Is this correct?
     
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  3. Aug 9, 2017 #2

    TSny

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    Yes, that's right.

    Yes, along with the fact that "the strength of magnetic field due to q2 at q1 i.e. B_2 is very small compared to B_1".
     
  4. Aug 15, 2017 at 12:50 AM #3
    Hello ,

    I can see how the direction of forces being exerted on each other are different .

    But how are you and the OP concluding that the strength of magnetic field B2 is very small compared to B1 ? Assuming charges are of equal magnitudes , I am not able to see how field strengths are being compared . Is their some implicit assumption ?
     
  5. Aug 15, 2017 at 1:12 AM #4

    TSny

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    The B-field of charge 1 at the location of charge 2 is much stronger than the field of charge 2 at the location of charge 1. In fact the B- field of 2 at the location of 1 is zero.

    See for example the following (which neglects relativistic effects)

     
  6. Aug 15, 2017 at 1:47 AM #5
    As depicted in the OP q1 is moving with speed v1 along +z and q2 is moving with speed v2 along +x axis .

    B1 (due to q1 ) = ##\frac{\mu}{4 \pi r^2}qv_1sin\theta_1 ##

    B2 (due to q2 ) = ##\frac{\mu}{4 \pi r^2}qv_2sin\theta_2 ##

    Which of the factors make B1 stronger than B2 ?
     
  7. Aug 15, 2017 at 2:11 AM #6
    Strictly speaking, you cannot use Biot - Savart law to calculate magnetic field due to a moving charged particle as the current is not steady.
    See the following in the Griffith's Electrodynamics book,
    upload_2017-8-15_12-31-31.png
    upload_2017-8-15_12-32-0.png
    upload_2017-8-15_12-32-35.png

    In case of calculating B2 at the position of q1 using your approach and considering griffith's footnote 11, ##\theta_ 2 ## ( which is angle between ## \vec v_2 ## and ## \vec r_1 - \vec r_2 ## i.e. position vector of q1 relative to q2) is ## \pi ##, which makes B2 zero at q1.
     
  8. Aug 15, 2017 at 10:36 AM #7
    Do you mind explaining this ?
     
  9. Aug 15, 2017 at 11:34 AM #8

    TSny

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    For a point charge moving at constant velocity, the B field circles around the line of motion of the charge. For the OP's question, the relevant facts are
    (1) B = 0 at all points on the line of motion of the charge.
    (2) B ≠ 0 at points that are not on the line of motion of the charge.

    A diagram of the field is shown here: https://www.miniphysics.com/uy1-magnetic-field-of-a-moving-charge.html

    Charge 2 does not produce any B field at the location of charge 1 because charge 1 lies on the line of motion of charge 2. However, charge 1 does produce some magnetic field at the location of particle 2 because 2 does not lie on the line of motion of 1.
     
  10. Aug 15, 2017 at 12:33 PM #9
    Oh ! I apologise . I didn't pay enough attention to the grey line connecting the two charges in OP's diagram . I was assuming the two charges moving with their instantaneous velocities along the axes but not in the way as shown in the OP .Sorry , again .

    By the way , two bar magnets exert equal and opposite forces obeying Newton's Third Law . So magnetic forces do obey Newton's Third Law . Isn't it ?
     
  11. Aug 15, 2017 at 2:05 PM #10

    TSny

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    Yes, forces between poles of magnets obey the 3rd law if the poles are at rest. Otherwise, maybe not. For example, if one of the poles is accelerating wildly, all bets are off.
     
  12. Aug 15, 2017 at 11:13 PM #11
    What you mean by this is there should be no relative motion between the magnets. Right?
     
  13. Aug 15, 2017 at 11:32 PM #12

    TSny

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