# Exact formula in simplest terms for Magnetic field at a point outside a solenoid [closed]

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)

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.

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

• 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. Dec 27, 2014 at 12:43
• 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 Dec 27, 2014 at 13:42