Faraday's Law of Induction — Question 2

Section 30-2 · Physics pages 4774–4775

Question 2 A spatial field gradient must be integrated before its time derivative is taken A stationary rectangular N-turn coil spans 0 < x < a and 0 < y < b. Its normal is +z and Bz(x, t) = βt(1 + x/a). Bz(x, t) out of page positive current: counterclockwise Schematic; labels, not drawing lengths, specify the data. (a) Find its flux linkage and induced emf. (b) Compare the exact emf with using the field at the left edge and at the midpoint.
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Question 2 - Solution A spatial field gradient must be integrated before its time derivative is taken Bz(x, t) out of page positive current: counterclockwise (a) Method and calculation Λ = N ∫b 0 dy ∫a 0 βt(1 + x/a)dx = 3 2Nabβt. E = −3 2Nabβ . (b) Interpretation and further calculation Eleft = −Nabβ = 2 3E. Bz(a/2, t) = 3 2βt ⇒Emidpoint = E . Physical check The midpoint works exactly for a linear spatial profile, not for arbitrary nonuniform fields.

Question 2

A spatial field gradient must be integrated before its time derivative is taken

A stationary rectangular N-turn coil spans 0<x<a0<x<a and 0<y<b0<y<b. Its normal is +z and Bz(x,t)=βt(1+x/a)B_z(x,t)=\beta t(1+x/a).

Diagram shown in the original SVG worksheet.

  1. Find its flux linkage and induced emf.

  2. Compare the exact emf with using the field at the left edge and at the midpoint.

Question 2 - Solution

A spatial field gradient must be integrated before its time derivative is taken

Diagram shown in the original SVG worksheet.

(a) Method and calculation

Λ=N∫0bdy∫0aβt(1+x/a)dx=32Nabβt.\Lambda=N\int_0^b\!dy\int_0^a\beta t(1+x/a)dx=\frac32Nab\beta t.ℰ=−32Nabβ.\boxed{\mathcal E=-\frac32Nab\beta}.

(b) Interpretation and further calculation

ℰleft=−Nabβ=23ℰ.\mathcal E_{\mathrm{left}}=-Nab\beta=\frac23\mathcal E.Bz(a/2,t)=32βt⇒ℰmidpoint=ℰ.B_z(a/2,t)=\frac32\beta t\Rightarrow\boxed{\mathcal E_{\mathrm{midpoint}}=\mathcal E}.

Physical check

The midpoint works exactly for a linear spatial profile, not for arbitrary nonuniform fields.