Mutual Induction — Question 10

Section 30-11 · Physics pages 4970–4971

Question 10 Sequential current ramps expose the mutual contribution to source work Two fixed lossless coupled coils have positive M. Starting from zero currents, first raise i1 to I1 while holding i2=0. Then hold i1=I1 and raise i2 to I2. Both final currents enter dotted terminals. L1 L2 i1 i2 positive currents enter dotted ends; M > 0 Schematic; labels, not drawing lengths, specify the data. (a) Compute the work supplied by each current source over the complete sequence. (b) Compare total work with magnetic energy and explain how reversing the ramp order changes individual source work.
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Question 10 - Solution Sequential current ramps expose the mutual contribution to source work L1 L2 i1 i2 positive currents enter dotted ends; M > 0 (a) Method and calculation W (a) 1 = ∫I1 0 L1i1di1 = 1 2L1I2 1, W (a) 2 = 0. W (b) 1 = ∫I2 0 MI1di2 = MI1I2, W (b) 2 = 1 2L2I2 2. (b) Interpretation and further calculation W1 + W2 = 1 2L1I2 1 + 1 2L2I2 2 + MI1I2 = Uf . W reversed 1 = 1 2L1I2 1, W reversed 2 = 1 2L2I2 2 + MI1I2 . Physical check The source already holding a nonzero current supplies the mutual term during the other ramp. Total work is independent of ramp order.

Question 10

Sequential current ramps expose the mutual contribution to source work

Two fixed lossless coupled coils have positive M. Starting from zero currents, first raise i1 to I1 while holding i2=0. Then hold i1=I1 and raise i2 to I2. Both final currents enter dotted terminals.

Diagram shown in the original SVG worksheet.

  1. Compute the work supplied by each current source over the complete sequence.

  2. Compare total work with magnetic energy and explain how reversing the ramp order changes individual source work.

Question 10 - Solution

Sequential current ramps expose the mutual contribution to source work

Diagram shown in the original SVG worksheet.

(a) Method and calculation

W1(a)=∫0I1L1i1di1=12L1I12,W2(a)=0.W_1^{(a)}=\int_0^{I_1}L_1i_1di_1=\tfrac12L_1I_1^2,\quad W_2^{(a)}=0.W1(b)=∫0I2MI1di2=MI1I2,W2(b)=12L2I22.W_1^{(b)}=\int_0^{I_2}MI_1di_2=MI_1I_2,\quad W_2^{(b)}=\tfrac12L_2I_2^2.

(b) Interpretation and further calculation

W1+W2=12L1I12+12L2I22+MI1I2=Uf.\boxed{W_1+W_2=\tfrac12L_1I_1^2+\tfrac12L_2I_2^2+MI_1I_2=U_f}.W1reversed=12L1I12,W2reversed=12L2I22+MI1I2.\boxed{W_1^{\mathrm{reversed}}=\tfrac12L_1I_1^2,\quad W_2^{\mathrm{reversed}}=\tfrac12L_2I_2^2+MI_1I_2}.

Physical check

The source already holding a nonzero current supplies the mutual term during the other ramp. Total work is independent of ramp order.