Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

Home exercise 4.1.2

Interactive: Biot–Savart Field Lines

Home exercise 4.1.2

Finite solenoid

A solenoid of $N$ turns, length $L$ and radius $b$ is wound tightly on the $z$ -axis between $z = 0$ and $z = L$ and carries a constant current $I$ . (a) Treat the slice between $z^{'}$ and $z^{'} + d z^{'}$ as a circular loop carrying the current $(N / L) I d z^{'}$ and use the result of Problem 4.1.2 to write $d B$ at the point $P (0, 0, z)$ on the axis. (b) Integrate over $0 \leq z^{'} \leq L$ to find $B (P)$ . Express the answer with the two angles $θ_{1}$ , $θ_{2}$ under which the ends of the solenoid are seen from P. (c) Show that at the centre of a long solenoid ( $L ≫ b$ ) the field tends to $μ_{0} N I / L$ , and that at either end it is half of that.

Adapted from: D. K. Cheng, Field and Wave Electromagnetics, 2nd ed., Problem P.6-7.

Try it yourself first.

Strategy

Superposition of loops along the axis: each slice is a loop of Problem 4.1.2 at the height $z^{'}$ , so $h \to z - z^{'}$ , and the on-axis fields of all the slices simply add. The integral is again that of Problem 4.1.1.

a) One slice

The winding has $n = N / L$ turns per metre, so the slice $d z^{'}$ carries the current $d I = n I d z^{'}$ in a loop of radius $b$ lying at $z^{'}$ . By symmetry (rotation about $z$ , P on the axis) its field at P is along $a_{z}$ , and Problem 4.1.2 with $h \to z - z^{'}$ gives

d B = \frac{μ _{0} b ^{2} n I d z ^{'}}{2 [ b ^{2} + ( z - z ^{'} ) ^{2} ] ^{3/2}} a_{z} .

b) Integrate over the winding

With $u = z^{'} - z$ ,

B_{z} = \frac{μ _{0} n I b ^{2}}{2} \int_{- z}^{L - z} \frac{d u}{( b ^{2} + u ^{2} ) ^{3/2}} = \frac{μ _{0} n I b ^{2}}{2} [\frac{u}{b ^{2} b ^{2} + u ^{2}}]_{- z}^{L - z} = \frac{μ _{0} n I}{2} [\frac{L - z}{b ^{2} + ( L - z ) ^{2}} + \frac{z}{b ^{2} + z ^{2}}] .

Let $θ_{1}$ and $θ_{2}$ be the angles between the $+ z$ direction and the lines from P to the rims of the two ends (Figure 4.1.27a): $cos θ_{1} = z / b^{2} + z^{2}$ towards the end $z^{'} = 0$ (negative if P is below it) and $cos θ_{2} = - (L - z) / b^{2} + (L - z)^{2}$ towards the end $z^{'} = L$ . Then

Answer

B (P) = \frac{μ _{0} N I}{2 L} (cos θ_{1} - cos θ_{2}) a_{z} [T]

valid for P inside or outside the winding.

c) Long solenoid

At the centre $z = L /2$ both terms equal $\frac{L /2}{b ^{2} + L ^{2} /4}$ , so

B_{z} (\frac{L}{2}) = \frac{μ _{0} N I}{L} \frac{L /2}{b ^{2} + L ^{2} /4} = \frac{μ _{0} N I}{L ^{2} + 4 b ^{2}} L ≫ b \frac{μ _{0} N I}{L} = μ_{0} n I .

At the end $z = 0$ the second term vanishes and $B_{z} (0) = \frac{μ _{0} n I}{2} \frac{L}{b ^{2} + L ^{2}} \to \frac{μ _{0} n I}{2}$ : half the central value, because only one half-space of turns contributes.

Answer

B_{centre} \to μ_{0} n I = \frac{μ _{0} N I}{L}, B_{end} \to \frac{μ _{0} n I}{2} (L ≫ b)

(a) B_z along the axis. — (a) $B_{z}$ along the axis.

$(b) Field lines of a solenoid with L=4b. Home exercise 4.1.2. (a) The axial field of a finite solenoid, in units of \mu_0nI: for L=10b it is 98\,\% of \mu_0nI at the centre and very nearly half of its long-solenoid value at each end, for L=2b it never reaches the plateau. (b) Computed field lines of a solenoid with L=4b (the dots are the turns, seen in cross-section): the field is nearly uniform inside, the lines leak out through the ends and close round the outside, and \nabla\cdot\vec B=0 holds line by line.$

Figure 4.1.27. (b) Field lines of a solenoid with

L = 4 b

. Home exercise 4.1.2. (a) The axial field of a finite solenoid, in units of

μ_{0} n I

: for

L = 10 b

it is

98 %

μ_{0} n I

at the centre and very nearly half of its long-solenoid value at each end, for

L = 2 b

it never reaches the plateau. (b) Computed field lines of a solenoid with

L = 4 b

(the dots are the turns, seen in cross-section): the field is nearly uniform inside, the lines leak out through the ends and close round the outside, and

\nabla \cdot B = 0

holds line by line.

With $L = 10 b$ the centre field is 98 % of $μ_{0} n I$ and the ends get about half of it. Shorten $L$ to see the plateau disappear.

InteractiveLoading simulation…

Check

Units: $μ_{0} n I$ is (H/m)(1/m)(A) $=$ T. Ampère's formula (Home exercise 4.1.7) gives the infinite-solenoid value $μ_{0} n I$ directly, and the finite formula shows how fast it is reached: for $L = 10 b$ the centre field is already $98 %$ of $μ_{0} n I$ , and still $85 %$ of it one radius in from an end (Figure 4.1.27a). Compare the plot of Problem 4.1.2: a single loop falls to half of its centre value at $h = 0.77 b$ , the solenoid only at its ends. The field-line picture (Figure 4.1.27b) shows where the missing half goes at an end: the lines fan out and return outside the winding.

Home exercise 4.1.2: Finite solenoid