Gauss' Law for Magnetic Fields — Question 4

Section 32-1 · Physics pages 5178–5179

Question 4 A linear three-dimensional field must have zero sum of normal gradients A proposed field is B = (ax, by, cz) in a rectangular region, where a, b, and c are constants with units T/m. Schematic; labels, not drawing lengths, specify the data. (a) Use a small box to find the constraint imposed by zero closed magnetic flux. (b) For a=.20 T/m and b=-.05 T/m, find c and the flux through a unit-area z-face at z=2.00 m with normal +z.
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Question 4 - Solution A linear three-dimensional field must have zero sum of normal gradients (a) Method and calculation Φx = a∆x∆y∆z, Φy = b∆x∆y∆z, Φz = c∆x∆y∆z. a + b + c = 0 . (b) Interpretation and further calculation c = −a −b = −.150 T/m, Bz(2) = −.300 T. Φ+z = −.300 Wb (A = 1.00 m2) . Physical check A nonzero gradient in one direction must be balanced by the other directions.

Question 4

A linear three-dimensional field must have zero sum of normal gradients

A proposed field is 𝑩=(ax,by,cz)\mathbf B=(ax,by,cz) in a rectangular region, where a, b, and c are constants with units T/m.

Diagram shown in the original SVG worksheet.

  1. Use a small box to find the constraint imposed by zero closed magnetic flux.

  2. For a=.20 T/m and b=-.05 T/m, find c and the flux through a unit-area z-face at z=2.00 m with normal +z.

Question 4 - Solution

A linear three-dimensional field must have zero sum of normal gradients

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Φx=aΔxΔyΔz,Φy=bΔxΔyΔz,Φz=cΔxΔyΔz.\Phi_x=a\Delta x\Delta y\Delta z,\quad\Phi_y=b\Delta x\Delta y\Delta z,\quad\Phi_z=c\Delta x\Delta y\Delta z.a+b+c=0.\boxed{a+b+c=0}.

(b) Interpretation and further calculation

c=−a−b=−.150T/m,Bz(2)=−.300T.c=-a-b=-.150\,T/m,\quad B_z(2)=-.300\,T.Φ+z=−.300Wb(A=1.00m2).\boxed{\Phi_{+z}=-.300\,Wb\quad(A=1.00\,m^2)}.

Physical check

A nonzero gradient in one direction must be balanced by the other directions.