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It's known that at the center of an electromagnet/solenoid the magnetic field $B$ is strongest there. At the pole/edge of the solenoid/ or electromagnet is the magnetic field $B$ strong? Let's assume $B$ = $1$ $Tesla$ at the center, how different is $B$ at the edges?

In general, is it simple to create a $1$ $Tesla$ magnetic field or higher? MRI's use field's above $4$ $Tesla$ yet it seems very complex and requires massive power & cooling.

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Provided that the solenoid is sufficiently long (i.e., radius$\ll$length), it should be about a half of the field at the center. One can see this by noting that the upper and lower halves of the solenoid contribute equally to the field at center. If you remove one of them, you only have one half of the original B field. –  higgsss Jan 29 at 0:07
@higgsss What if the solenoid was not that long? Quite short in length and large in radius. It seems that it might be simple and possible to create a powerful electromagnet and the poles are near a $Tesla$ . –  Key Jan 29 at 7:19
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1 Answer 1

The magnetic field of a circular current thread lying in the y,z-plane with center at the origin is given as $B = \frac{\mu I}{2} \frac{R^2}{(x^2+R^2)^{\frac{3}{2}}}$ on the x-axis at the coordinate value $x$. The B-field is aligned to the x-axis. The direction of rotation of the current flow and the direction of the B-field build a right-handed screw.

We integrate this to find the B-field on the x-axis at the coordinate $x$ for a solenoid with $n$ windings of current $I$, and with the x-axis as its symmetry axis. The current flow is modelled as uniform current distribution of strength $\frac{nI}{L}$ on a cylinder surface with radius $R$ starting at $x_{\rm coil}=0$ and ending at $L$. The result of the integration is: $$B = \frac{\mu n I}{2L} \left[\frac{x}{\sqrt{x^2+R^2}} + \frac{(L-x)}{\sqrt{(L-x)^2+R^2}}\right].$$

Now at the ends the field is simply $$B_{\rm end} = \frac{\mu n I}{2\sqrt{L^2+R^2}}$$

At the middle of the coil the field on the x-axis is $$ B_{\rm middle} = \frac{\mu n I}{2\sqrt{\left(\frac L2\right)^2+R^2}} $$ Note : There is no change in material (permeability)
Here $\mu = \mu_r \mu_0 = k \mu_0$ and $n$ is no of turns per unit length.

In an MRI scanner the fields are produced around objects which are not in contact ! i.e. The fields have strengths ranging from 1 to 7 T in air, it is not very difficult to produce these fields in material like iron as they boost magnetic fields by factors of 1000. But to make such fields in air 1000 times energy is required and therefore these scanners utilise high power, coolants and nowadays superconducting wires too.

Addendum :
For an example (as requested in comments), suppose a solenoid of length $1 m$ and radius of cross section $0.1m$, having $1000$ no of turns per unit length($1 m$) carrying current $1 A$ and filled with air which makes $\mu = \mu_0$. Then, $$B = \frac{1000\mu_0}{2} \left[\frac{x}{\sqrt{x^2+0.1^2}} + \frac{(1-x)}{\sqrt{(1-x)^2+0.1^2}}\right].$$ $$B_{\rm end} = \frac{1000\mu_0}{2\sqrt{1^2+0.1^2}}$$ $$B_{\rm end} = 0.0025008012 T$$ $$B_{\rm middle} = \frac{1000\mu_0}{2\sqrt{\left(\frac{1}{2}\right)^2+0.1^2}}$$ $$B_{\rm middle} = 0.0049289361 T$$

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Note, iron saturates above 2T. This makes 7T difficult. –  Tobias Jan 30 at 0:46
@Tobias : I have not mentioned fields of 7T in iron, I have mentioned the upper limit of today's mri scanner field. I haven't talked about how it is achieved ! Also the formulae were correct before too as at that time $n$ was loops per unit length and now it is total no. of loops, thanks anyway for the edit ! –  Rijul Gupta Jan 30 at 9:52
@rijulgupta Could you please use an example to illustrate this, I did a few calculations but not sure of my values are correct... But could you use k in the formula of B.end? –  Key Feb 5 at 1:05
amazing! thank you @rijulgupta! –  Key Feb 5 at 16:05
@rijulgupta L = The length of the whole wire used to create the total number of turns correct? So 1 meter for that wire is used to created n # of turns? –  Key Feb 6 at 2:58
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