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Thread: Simple equation

  1. #1
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    Simple equation

    I would like to get q and m, such that $q > 1$, $m > 1$ and $B > 1$ and $$\frac{qB}{m} = 1$$, now q and m are the mass and charge of electron which is $10^{-27}$ something.

    So I multiply the numerator and denominator with same number, but in the numerator it should be $\sqrt{a}$ for $B$ and also for $q$, separately, right?
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  2. #2
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    Re: Simple equation

    wut?
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  3. #3
    Forum Admin topsquark's Avatar
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    Re: Simple equation

    Quote Originally Posted by Nforce View Post
    I would like to get q and m, such that $q > 1$, $m > 1$ and $B > 1$ and $$\frac{qB}{m} = 1$$, now q and m are the mass and charge of electron which is $10^{-27}$ something.

    So I multiply the numerator and denominator with same number, but in the numerator it should be $\sqrt{a}$ for $B$ and also for $q$, separately, right?
    There are two problems here.
    1) You can actually solve for B, which I'm sure isn't what you want.

    $\displaystyle q = 1.602 \times 10^{-19} \text{ C}$
    $\displaystyle m_e = 9.11 \times 10^{-31} \text{ kg}$

    So
    $\displaystyle \dfrac{q}{m_e} \approx 1.76 \times 10^{11} \text{ C/kg}$

    thus
    $\displaystyle \dfrac{qB}{m_e} = (1.76 \times 10^{11}) B = 1$

    and finally:
    $\displaystyle B = 5.69 \times 10^{-12} \text{ kg/C}$

    2) This isn't the correct unit for the magnetic induction, which is in T (for Tesla.) The unit you have is $\displaystyle kg/C$ and the unit for a tesla is $\displaystyle T ~ "="~ kg/sC$.

    And finally: What is $\displaystyle \sqrt{\alpha}$? I don't recognize your equation of the top of my head so I don't know what you are trying to talk about.

    -Dan
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    Re: Simple equation

    Ok, here is the problem. I am numerically solving equations in Matlab. If I use the actual value of mass and charge of the electron then the time complexity gets higher and I am not going to wait 2 days to solve an equation. So I would like to multiply the fraction with some term let's call it $a$ so that I would get a higher number for mass and charge. Something above 1.

    So not to change the fraction I would multiply the numerator and denominator with the same number $\frac{a}{a}$. But because we have $B$ and $q$ in the numerator I would multiply like this:

    $$\frac{\sqrt{a}q \sqrt{a}B }{a m} $$

    Is this then the correct way?

    Thanks.
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    Re: Simple equation

    Quote Originally Posted by Nforce View Post
    I would like to get q and m, such that $q > 1$, $m > 1$ and $B > 1$ and $$\frac{qB}{m} = 1$$, now q and m are the mass and charge of electron which is $10^{-27}$ something.

    So I multiply the numerator and denominator with same number, but in the numerator it should be $\sqrt{a}$ for $B$ and also for $q$, separately, right?
    There is no "a" in your original $\frac{qB}{m}= 1$. What does "a" refer to?
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    Re: Simple equation

    Quote Originally Posted by HallsofIvy View Post
    There is no "a" in your original $\frac{qB}{m}= 1$. What does "a" refer to?
    See my post #4. The "a" is a factor which I multiply both the numerator and denominator.
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  7. #7
    Forum Admin topsquark's Avatar
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    Re: Simple equation

    Quote Originally Posted by Nforce View Post
    See my post #4. The "a" is a factor which I multiply both the numerator and denominator.
    Yes but you haven't explained why you would do that. I see no value in this approach as it changes nothing.

    -Dan
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    Re: Simple equation

    Because I would like to keep the ratios between physical quantities the same. The reason is that very small values of charge and mass, which are near zero increases time complexity. So I would like to have them above 1.
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    Re: Simple equation

    Do you understand what am I trying to say? Is this the right way?
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    Forum Admin topsquark's Avatar
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    Re: Simple equation

    Quote Originally Posted by Nforce View Post
    Do you understand what am I trying to say? Is this the right way?
    It should be doable. But since you can get a value for B directly why would you want to do an approximation? Or are you just playing around with it?

    -Dan
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