ch30-p042 - 42. (a) We imagine dividing the one-turn...

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Unformatted text preview: 42. (a) We imagine dividing the one-turn solenoid into N small circular loops placed along the width W of the copper strip. Each loop carries a current ∆i = i/N. Then the magnetic field inside the solenoid is −7 ⎛ N ⎞ ⎛ i ⎞ µ i (4π×10 T ⋅ m/A)(0.035A) = 2.7 ×10−7 T. B = µ0 n∆i = µ0 ⎜ ⎟ ⎜ ⎟ = 0 = 0.16m ⎝W ⎠⎝ N ⎠ W (b) Eq. 30-33 leads to L= 2 Φ B πR 2 B πR ( µ 0i / W ) πµ 0 R 2 π (4π×10−7 T ⋅ m/A)(0.018m) 2 = = = = = 8.0 ×10−9 H. i i i W 0.16m ...
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