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Initial Conditions

A charged particle $q$, is initially at rest at $(x,y)$. It is given velocity, $v_o~m.s^{-1}$ along positive $X$-axis.

There exists a Uniform Magnetic Field $B_0$ in the whole region along positive $Z$-axis.

After the particle in given the velocity, we know that it experiences some force, due to the presence of Magnetic Field, in a direction perpendicular to both $B$ & $v$. This causes the charged particle to move in a cirle of radius, $r = \frac{m.v}{q.B}$

So, how do we find the centre of this circular motion ?

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  • $\begingroup$ Just using the right hand rule and that charge comes in two polarities you can show there are two circles that can be made by the particle. $\endgroup$
    – Triatticus
    Commented Apr 20, 2021 at 16:43
  • $\begingroup$ I meant to ask about only for one type of charge [either positive or negative], so I have edited the question to be specific about positive charge now $\endgroup$
    – gobbledy-gook
    Commented Apr 21, 2021 at 1:06
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    $\begingroup$ The center depends on the initial conditions. The initial position and velocity will be on a tangent to the circle. $\endgroup$
    – sasquires
    Commented Apr 21, 2021 at 1:46
  • $\begingroup$ If the charge and magnetic field are constant you can also get an infinite number of circular paths from the velocity alone $\endgroup$
    – Triatticus
    Commented Apr 21, 2021 at 5:18
  • $\begingroup$ If started from origin with, let us say, v = i + j m/s then there will be one circle or more than one, this is what I wanted to know $\endgroup$
    – gobbledy-gook
    Commented Apr 21, 2021 at 7:54

1 Answer 1

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In a right handed system with, B, in the, +z , direction, and, v, at the origin, and going in the, +x, direction, a positive charge will be deflected toward the, -y, direction. The center of the circle will be on the y axis at: y = -r.

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  • $\begingroup$ So, you mean that (0,-r) will be center but why? why not (0,0) ? could you please give more details on it $\endgroup$
    – gobbledy-gook
    Commented Apr 21, 2021 at 15:49
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    $\begingroup$ (0,0) is one of the points on the circle, (the charge starts there) so it can't be the center. $\endgroup$
    – Math Keeps Me Busy
    Commented Apr 23, 2021 at 5:52

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