Magnetic Field of a Coil

Using the Biot-Savart law, we find that the field on the axis of a circular current loop is B = m_oiR²/2(x²+R²)^3/2, where x is the distance to the center of the loop. For N turns, simply multiply by N if they are all about the same distance x, i.e., a very short coil. If the windings are spread out over length L, consider dn windings in length dx: dn = (N/L)dx. Then integrate. This is easiest if you convert to q as the variable; see below.

q

p

The rectangle represents the coil and point p on the axis is the point where we wish to find the magnetic field. P could be outside the coil as well. Note that
sinq = R/(x²+R²)^1/2, so B = m_oi sin³q/2R. Also x = Rcotq, so dx = -Rcsc²qdq. Plug these in and we find that

B =-(m_oNi/2L) òsinqdq

The result has the difference of two cosines, one of which is in the 2^nd quadrant and is therefore negative. In terms of the first quadrant angles shown below,

a b
B = (m_oNi/2L)(cosa+cosb)

If point p is to the right of the coil above, b exceeds 90^o, and cosb is negative, so we go back to the difference between two first quadrant cosines. At a large distance x from the center of the coil, on the coil’s axis, replace this difference of cosines with sinqdq, where sinq » tanq = R/x and dq = R/(x-L/2) – R/(x+L/2). Do some algebra on this and get
B = m_oNiR²/[2x(x²-L²/4)] or if L<<x, B = m_oNiR²/2x³

Since the face area is pR², the above is B= m_oNiA/[2px(x²-L²/4)] or B = m_oNiA/2px³. NiA is called the magnetic moment

For a long coil, B at one end has one cosine » 1 and the other = 0. Call B at the end B_o, and we find that it is m_oNi/2L. Using this, see that at large x on the axis outside the coil, B = B_oLR²/[x(x²-L²/4)] =B_oLA/[px(x²-L²/4)]. The same result should be true for a cylindrical permanent magnet, and if the magnet has a rectangular cross section, use the formula with the A in it

My main pages: on mechanics fluids, heat, e&m, vibrations& waves, quantum, index. fredrick.gram@ tri-c.edu

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