Kvant Physics Problem 1000
A cube of edge length $a$ carries a steady current $I$ along a closed contour formed by its edges.
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Problem
Current $I$ flowing along the contour formed by four edges of a cube (Fig. 2) creates at the center of the cube a magnetic field with induction $B_0$. Find the magnitude and direction of the magnetic induction vector at the center of the cube produced by a current $I$ flowing along a contour consisting of six edges (Fig. 3).
Figure 2 Figure 3
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M. M. Tsypin
Setup and Assumptions
A cube of edge length $a$ carries a steady current $I$ along a closed contour formed by its edges. The magnetic induction is required at the cube center $O$.
In the first configuration (Fig. 2), the current flows along four edges forming one square face of the cube. The magnetic induction at $O$ produced by this contour has magnitude $B_0$.
In the second configuration (Fig. 3), the current flows along six edges of the cube forming a closed contour that coincides with the boundary of three mutually adjacent faces meeting at one vertex of the cube.
The magnetic field is evaluated in vacuum, and the current distribution is treated as thin wires carrying steady current. Superposition of magnetic fields from separate wire segments is valid. Edge effects other than geometry are neglected.
Physical Principles
The magnetic field of a steady current distribution is given by the Biot–Savart law for a thin wire element,
$$d\mathbf{B} = \frac{\mu_0 I}{4\pi} \frac{d\mathbf{l} \times \mathbf{r}}{r^3}.$$
The total magnetic field is the vector sum of contributions from all wire segments.
For a set of wire contours forming boundaries of surfaces, the field at a point can be obtained by decomposing the contour into simpler closed loops, using linearity of the Biot–Savart law.
Symmetry implies that for a square loop, the magnetic field at a point on its symmetry axis is directed along the axis normal to the loop.
Derivation
Let the cube have center at the origin and edges aligned with the coordinate axes. Consider first a single square face of the cube carrying current $I$ along its four edges. This loop lies in a plane, for example $z = a/2$.
At the cube center, which lies on the axis perpendicular to this face through its center, the magnetic field produced by this loop is directed along the normal to the face. Its magnitude is given as $B_0$.
By cubic symmetry, the same result holds for any face of the cube: a square loop forming any face produces at the center a magnetic field of magnitude $B_0$ directed along the outward normal to that face.
The six-edge contour in Fig. 3 can be represented as the boundary of three mutually perpendicular square faces that meet at one vertex of the cube. These three faces can be chosen as the faces perpendicular to the $x$-, $y$-, and $z$-axes that share a common corner.
The boundary of this union of three faces coincides exactly with the six-edge contour, with consistent orientation of current along the edges.
The magnetic field at the center produced by this contour equals the vector sum of the fields produced by the three individual face loops, since contributions from shared edges cancel when the surface is decomposed into oriented boundaries.
Each face loop produces a field of magnitude $B_0$ at the center, directed along its outward normal. Choosing the orientation consistent with the contour gives three mutually perpendicular contributions,
$$\mathbf{B}_x = B_0 \hat{\mathbf{x}}, \quad \mathbf{B}_y = B_0 \hat{\mathbf{y}}, \quad \mathbf{B}_z = B_0 \hat{\mathbf{z}}.$$
The total magnetic induction at the center is
$$\mathbf{B} = \mathbf{B}_x + \mathbf{B}_y + \mathbf{B}_z = B_0(\hat{\mathbf{x}} + \hat{\mathbf{y}} + \hat{\mathbf{z}}).$$
Its magnitude follows from orthogonality of unit vectors,
$$B = B_0 \sqrt{1 + 1 + 1} = \sqrt{3},B_0.$$
The direction is along the space diagonal of the cube passing through the center, corresponding to the vector $\hat{\mathbf{x}} + \hat{\mathbf{y}} + \hat{\mathbf{z}}$, i.e., toward the vertex common to the three faces used in the contour orientation.
Result
The magnetic induction at the cube center for the six-edge contour is
$$\mathbf{B} = B_0(\hat{\mathbf{x}} + \hat{\mathbf{y}} + \hat{\mathbf{z}}).$$
Its magnitude is
$$B = \sqrt{3},B_0.$$
No additional numerical substitution is required because the result is expressed in terms of the given reference field $B_0$.
Sanity Checks
Each face-loop contribution has units of tesla, so $\mathbf{B}$ retains units of tesla, and the factor $\sqrt{3}$ is dimensionless.
The result scales linearly with current $I$ because each face contribution is proportional to $I$, consistent with the Biot–Savart law.
If the three face contributions were pairwise opposite, symmetry would produce cancellation and $B=0$, but the geometry selects three orthogonal, not opposing, contributions, so cancellation does not occur.
Replacing one face loop by the known configuration recovers $B=B_0$, consistent with the definition of the reference field.