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I need the COMPLETE formula for magnetic field outside the solenoid.

So the situation I am stuck in is I have to solve this question:

The magnetic field at the centre of coil of $n$ turns, bent in the form of a square of side $2l$, carrying current $I$ is

Options: (A) $\frac {\sqrt{2}μ_0nI}{\pi l}$ (B) $\frac {\sqrt{2}μ_0nI}{2\pi l}$ (C) $\frac {\sqrt{2}μ_0nI}{4\pi l}$ (D)$\frac {2μ_0nI}{\pi l}$

This is how I visualized it:

(I am really, really sorry for the poor drawing, but I could figure out a better software)

enter image description here

SO I think of this as 4 circular coils/solenoids (and I arbitrarily took the direction of current, since I am not asked to find direction of magnetic field, only magnitude, this shouldn't matter), and I see that solenoids on opposite sides of the square have same direction of magnetic field.

Now I think of the formula for the magnetic field at a point outside the solenoid/circular coil as

$B = \frac{μ_0}{4\pi} \times \frac {NI}{R}$

where $N$ is number of turns, $I$ is current through solenoid/circular coil and $R$ is perpendicular distance of point from the circular coil/solenoid.

So when I apply this formula for each side of the square, I take $N = n/4$ and $R = l$

$B = \frac {μ_0nI}{4\pi4l}= \frac {μ_0nI}{16\pi l} $

Since opposite sides have same direction of magnetic field, the resultant magnetic field is the magnitude of the vector sum of perpendicular vectors having magnitude $2B$.

Since they are perpendicular: the magnitude is just

$\sqrt {(2B)^2 + (2B)^2}= \sqrt {2 \times (2B)^2 } = \sqrt{2} \times 2B = \sqrt{2} \times \frac {μ_0nI}{8\pi l} = \frac {\sqrt{2}μ_0nI}{8\pi l}$

Now this is close to the options, but I fear that I am missing something in my formula for the magnetic field. Hence I request you to correct my formula.

NOTE: I only need the formula, please give me the formula, and not means to derive it from Biot-Savart and Ampere's circuital law. Derivations I will be doing 2 years later when I have Biot-Savart and Amprer's circuital law in my curriculum.

NOTE 2: You could call that a duplicate of my question https://physics.stackexchange.com/questions/155119/a-few-questions-related-to-magnetic-fields but I request not to close this as that has been put on hold and I am sure its visibility is damaged already and I can't wait for busy moderators to open the question again.

Since, I only need the formula (assuming that my approach to the question is correct), I request anyone with sufficient knowledge to either post the formula or correct my approach ASAP.

Thanks in advance.

PLEASE, PLEASE, PLEASE do NOT CLOSE this question straightforwardly if you have any objections, tell me what's amiss in the comments.

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1 Answer 1

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I think you brought up solenoid unnecessarily, look up Magnetic field at center of a square and then apply it for coil having n turns. It would look something like this

I will link you the derivation hoping you understand something from it.

Magnetic field at center of square loop is $\frac{\sqrt{2} \mu_{0} I}{\pi l}$

and for n turns is $\frac{\sqrt{2} n\mu_{0} I}{\pi l}$

Reg other questions you asked i have given my view on this .

  1. When Helium nucleus makes a full rotation. We may consider it as current in loop so magnetic field at the centre of loop

$$\frac{\mu_{0} nI}{2R}$$ n=1 in this case, $I=\frac{q}{t}$ = $\frac{2e}{2}$ =e(charge of an electron)

$$B= \frac{\mu_{0} nI}{2R}= \frac{\mu_{0} e}{2R} = \frac{\mu_{0} e}{2R} =\frac{\mu_{0} 1.6 * 10^{-19}}{2*0.8} = {\mathbf{\mu_{0} 10^{-19}}} $$

  1. The force on a charge moving in a uniform magnetic field is given by , $F_{b}=q(\vec{v}\times\vec{B})$ . If the charge enters with a $\vec{v}$ perpendicular to the $\vec{B}$ then it will move in a circle of radius r. Centripetal force on a particle is $F= \frac{mv^{2}}{r}$ Equating these two , we will get $$ \frac{mv^{2}}{r} = qvB \implies p (mv)= Bqr -->1 $$
    p-> momentum

Kinetic energy of a particle = $\frac{1}{2}mv^{2}=\frac{m^{2}v^{2}}{2m} \implies K.E= \frac{p^{2}}{2m}\implies p=\sqrt{2mK.E} -->2$

equating 1 and 2 , $Bqr =\sqrt{2mK.E} \implies r=\frac{\sqrt{2mK.E}}{Bq} \implies r \propto \frac{\sqrt{m}}{q} $

$m_{\alpha} = 4m_{p} ; m_{D}=2m_{p}; q_{\alpha} = 2q_{p} ; q_{D}=q_{p} $

From this $\mathbf{r_{D}>r_{\alpha}=r_{p}}$ Ans : Option A

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  • $\begingroup$ Doubt: As per Biot-Savart law, for any current carrying conductor, $dB = \frac {mu_0}{4\pi} \times \frac {Idl\sin\theta}{r^2} $ ($r$ is distance of point from conductor, $dl$ is infinitesimal portion of the length of the conductor, $dB$ is magnetic field created by that infinitesimal part). But $\sin\theta = 1$ only in one case, when the part in conductor is directly perpendicular to the point otherwise $\sin\theta$ varies. B = $dB_a + dB_b + ..$, where $dB_x$ is magnetic field by point x. But they are not equal as $\sin\theta$ varies. Yet replacing dB,dl with B,l works for entire conductor. $\endgroup$
    – Swapnil Rustagi
    Dec 27, 2014 at 12:43
  • $\begingroup$ i have included the derivation , We take the angle in the integration which we perform to find the field. If you take a look at that and still have doubt comment, i will reply . Also i included solution to your other 2 questions $\endgroup$
    – Gowtham
    Dec 27, 2014 at 13:42

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