Homework 12 Solutions

# 32 18 and eq 32 19 equal to 40 cm20 cm enclosed area

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Unformatted text preview: 10−7 T ⋅ m A 8.85 ×10−12 C2 /N ⋅ m 2 3.0 × 10−3 m ( )( ) 2 = 2.4 × 1013 V . m ⋅s 6. The integral of the field along the indicated path is, by Eq. 32-18 and Eq. 32-19, equal to (4.0 cm)(2.0 cm) ⎛ enclosed area ⎞ μ0id ⎜ = 52 nT ⋅ m . ⎟ = μ0 (0.75 A) 12 cm 2 ⎝ total area ⎠ 7. (a) Inside we have (by Eq. 32-16) B = μ0id r1 / 2π R 2 , where r1 = 0.0200 m, R = 0.0300 m, and the displacement current is given by Eq. 32-38 (in SI units): id = ε 0 dΦE = (8.85 × 10−12 C2 /N ⋅ m 2 )(3.00 × 10−3 V/m ⋅ s) = 2.66 × 10−14 A . dt Thus we find μ0id r1 (4π ×10−7 T ⋅ m/A)(2.66 × 10−14 A)(0.0200 m) B= = = 1.18 × 10−19 T . 2 2 2π R 2π (0.0300 m) (b) Outside we have (by Eq. 32-17) B = μ0id / 2π r2 where r2 = 0.0500 cm. Here we obtain μ0id (4π ×10−7 T ⋅ m/A)(2.66 ×10−14 A) = = 1.06 ×10−19 T B= 2π r2 2π (0.0500 m) 8. (a) Application of Eq. 32-3 along the circle referred to in the second sentence of the problem statement (and taking the derivative of the flux expression given in that sentence) leads to r B (2π r ) = ε 0 μ0 ( 0.60 V ⋅ m/s ) . R Using r = 0.0200 m (which, in any case, cancels out) and R = 0.0300 m, we obtain 1245 B= ε 0 μ0 (0.60 V ⋅ m/s) (8.85 × 10−12 C2 /N ⋅ m 2 )(4π×10−7 T ⋅ m/A)(0.60 V ⋅ m/s) = 2π R 2π (0.0300 m) = 3.54 × 10−17 T . (b) For a value of r larger than R, we must note that the flux enclosed has already reached its full amount (when r = R in the given flux expression). Referring to the equation we wrote in our solution of part (a), this means that the final fraction ( r / R ) should be replaced with unity. On the left hand side of that equation, we set r = 0.0500 m and solve. We now find B= ε 0 μ0 (0.60 V ⋅ m/s) (8.85 ×10−12 C2 /N ⋅ m 2 )(4π×10−7 T ⋅ m/A)(0.60 V ⋅ m/s) = 2π r 2π (0.0500 m) = 2.13 × 10−17 T . 9. (a) Application of Eq. 32-7 with A = πr2 (and taking the derivative of the field expression given in the problem) leads to B (2π r )...
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## This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

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