Gauss' Law for Magnetic Fields — Question 1

Section 32-1 · Physics pages 5172–5173

Question 1 One missing face flux is fixed by the five measured outward fluxes A closed box has outward magnetic fluxes +4.0, -2.0, +1.5, -3.0, and +.5 mWb through five faces. Its sixth face has area .0200 square meters. Schematic; labels, not drawing lengths, specify the data. (a) Find the sixth outward flux. (b) Find its mean outward normal field and explain whether zero total flux requires zero field everywhere.
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Question 1 - Solution One missing face flux is fixed by the five measured outward fluxes (a) Method and calculation I B · dA = 0 ⇒Φ6 = − 5 X j=1 Φj. Φ6 = −(4 −2 + 1.5 −3 + .5) = −1.00 mWb . (b) Interpretation and further calculation ⟨Bn⟩= Φ6/A6 = −.00100/.0200. ⟨Bn⟩= −.0500 T; I B · dA = 0̸ ⇒B = 0 . Physical check The minus sign denotes net entry through that face, with outward normals used consistently.

Question 1

One missing face flux is fixed by the five measured outward fluxes

A closed box has outward magnetic fluxes +4.0, -2.0, +1.5, -3.0, and +.5 mWb through five faces. Its sixth face has area .0200 square meters.

Diagram shown in the original SVG worksheet.

  1. Find the sixth outward flux.

  2. Find its mean outward normal field and explain whether zero total flux requires zero field everywhere.

Question 1 - Solution

One missing face flux is fixed by the five measured outward fluxes

Diagram shown in the original SVG worksheet.

(a) Method and calculation

∮𝑩⋅d𝑨=0⇒Φ6=−∑j=15Φj.\oint\mathbf B\cdot d\mathbf A=0\Rightarrow\Phi_6=-\sum_{j=1}^5\Phi_j.Φ6=−(4−2+1.5−3+.5)=−1.00mWb.\boxed{\Phi_6=-(4-2+1.5-3+.5)=-1.00\,mWb}.

(b) Interpretation and further calculation

⟨Bn⟩=Φ6/A6=−.00100/.0200.\langle B_n\rangle=\Phi_6/A_6=-.00100/.0200.⟨Bn⟩=−.0500T;∮𝑩⋅d𝑨=0⇏𝑩=0.\boxed{\langle B_n\rangle=-.0500\,T;\quad\oint\mathbf B\cdot d\mathbf A=0\not\Rightarrow\mathbf B=0}.

Physical check

The minus sign denotes net entry through that face, with outward normals used consistently.