Irodov 3.241: Half the Flux Leaves Before the End – Amit & Deepali Sinha Foundation

Problem Irodov 3.241

A straight solenoid, so long that it may be treated as infinite in one direction, has $n$ turns per unit length, a cross-sectional area $S$ , and carries a current $I$ . What is the flux of the magnetic induction $\vect B$ through the flat cross-section at the solenoid’s end?

The end of the very long winding, which runs on to the left: n turns per unit length carry the current I, and the cross-section has area S. — The end of the very long winding, which runs on to the left: $n$ turns per unit length carry the current $I$ , and the cross-section has area $S$ .

Given

$n$ turns per unit length
$S$
$I$

Find

\Phi

through the end cross-section

Solution

Step 1The field and the flux deep inside

Far from the end the solenoid is effectively infinite, and Ampere’s law gives the familiar uniform field along the axis.

B_0 = \mu_0 n I \tag{1}

Through a cross-section there, the flux is therefore

\Phi_0 = B_0 S = \mu_0 n I S \tag{2}

At the end the field is no longer uniform, so a direct integration over the end face would need the full off-axis field. A symmetry argument avoids it.

Step 2Two half solenoids make an infinite one

Treat the winding as a sheet of azimuthal current $nI$ per unit length, the pitch of the turns being negligible. Place a second identical semi-infinite solenoid against the end, wound the same way and carrying the same current, so that the pair forms one infinite solenoid. Call the given half $1$ and the added half $2$ ; on the joint plane, which is the end plane of half $1$ , superposition gives the axial components (along $z$ , the axis, pointing out of the end), with $r$ the distance from the axis:

B_{1z} + B_{2z} = \begin{cases} B_0, & r < R\\ 0, & r > R\end{cases} \tag{3}

where $R$ is the radius of the winding, so that $S = \pi R^2$ .

Step 3The mirror in the end plane makes the halves equal

Reflect half $1$ in the end plane. The image is half $2$ : the current sheet maps onto the sheet at the mirror position, and its azimuthal current keeps its sense, because the reflection reverses only the axial direction. The field transforms with it, and at a point on the mirror plane the axial component, which is normal to that plane, is unchanged, so at every point of the joint plane

B_{2z} = B_{1z}

Substituting this into (3):

2B_{1z} = \begin{cases} B_0, & r < R\\ 0, & r > R\end{cases}

B_{1z} = \frac{B_0}{2} \quad\text{at every point of the end face}

(4)

Step 4The flux through the end face

The end of the solenoid cut along its axis (dots and crosses: the current I in the turns), with the computed B lines seeded by equal flux deep inside: half of them cross the end plane S, the other half leave between the turns before it. — The end of the solenoid cut along its axis (dots and crosses: the current $I$ in the turns), with the computed $\vect B$ lines seeded by equal flux deep inside: half of them cross the end plane $S$ , the other half leave between the turns before it.

The axial component (4) is the same over the whole face, so the flux is that value times the area; with (1) and (2):

\Phi = \frac{B_0}{2}\,S

\Phi = \frac{\Phi_0}{2} = \frac{\mu_0 n I S}{2}

(5)

Check

On the axis alone, a solenoid’s field is $(\mu_0 nI/2)(\cos\theta_1 - \cos\theta_2)$ , with $\theta_1$ and $\theta_2$ the angles at which the two ends of the winding are seen from the field point. At the end of a semi-infinite solenoid one end is seen at $90^\circ$ and the other at $180^\circ$ , so the bracket is $0 - (-1) = 1$ and $B = \mu_0 nI/2$ , the value (4) gives there. Dimensionally, $\mu_0 nI$ is a field in tesla and $S$ an area, so $\Phi$ is in webers.

Remark

The symmetry gives more than the flux: the axial field is exactly $B_0/2$ at every point of the end face, not just on average, and exactly zero on the end plane outside the winding. Since $\nabla\cdot\vect B = 0$ , the other half of $\Phi_0$ must leave through the side of the winding before the end, which is what the field lines of the last figure show; none of the returning flux crosses the end plane outside the solenoid.