Homework 12 Solutions

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 )...
View Full Document

This document was uploaded on 02/26/2014 for the course PHYS 2b at UCSD.

Ask a homework question - tutors are online