Gauss' Law for Magnetic Fields — Question 5

Section 32-1 · Physics pages 5180–5181

Question 5 Solenoidality reconstructs a missing transverse component In a region, Bx = αx2y, Bz = 0, and By(x, 0, z) = 0. The constant alpha carries the units required for B to be in teslas. Schematic; labels, not drawing lengths, specify the data. (a) Find By using ∇· B = 0. (b) Verify the divergence of the reconstructed field and state why the boundary value is needed.
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Question 5 - Solution Solenoidality reconstructs a missing transverse component (a) Method and calculation ∂xBx = 2αxy ⇒∂yBy = −2αxy. By = −αxy2 + f(x, z), By(x, 0, z) = 0 ⇒By = −αxy2 . (b) Interpretation and further calculation ∇· B = 2αxy −2αxy + 0 = 0. Without boundary data, any f(x, z) remains undetermined. Physical check The divergence equation constrains a derivative, so an integration function survives until boundary data fix it.

Question 5

Solenoidality reconstructs a missing transverse component

In a region, Bx=αx2yB_x=\alpha x^2y, Bz=0B_z=0, and By(x,0,z)=0B_y(x,0,z)=0. The constant alpha carries the units required for B to be in teslas.

Diagram shown in the original SVG worksheet.

  1. Find By using ∇⋅𝑩=0\nabla\cdot\mathbf B=0.

  2. Verify the divergence of the reconstructed field and state why the boundary value is needed.

Question 5 - Solution

Solenoidality reconstructs a missing transverse component

Diagram shown in the original SVG worksheet.

(a) Method and calculation

∂xBx=2αxy⇒∂yBy=−2αxy.\partial_xB_x=2\alpha xy\Rightarrow\partial_yB_y=-2\alpha xy.By=−αxy2+f(x,z),By(x,0,z)=0⇒By=−αxy2.B_y=-\alpha xy^2+f(x,z),\quad B_y(x,0,z)=0\Rightarrow\boxed{B_y=-\alpha xy^2}.

(b) Interpretation and further calculation

∇⋅𝑩=2αxy−2αxy+0=0.\nabla\cdot\mathbf B=2\alpha xy-2\alpha xy+0=0.Without boundary data, any f(x,z) remains undetermined.\boxed{\text{Without boundary data, any }f(x,z)\text{ remains undetermined.}}

Physical check

The divergence equation constrains a derivative, so an integration function survives until boundary data fix it.