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Unformatted text preview: PHYSICS 218 SOLUTION TO HW 3 Created: September 12, 2004 10:42pm Last updated: October 22, 2004 3:06pm 1. (a) Instead of looking up the units in a table directly, we will argue from what we know. As . 5 CV 2 is the energy stored in a capacitance we can deduce [ C ] = F = J/V 2 and similarly as . 5 LI 2 is the energy of the magnetic field due to a steady current we have [ L ] = H = J/A 2 . Putting this together we find r 1 L 1 C 1 = m r V 2 A 2 J 2 = m s (1) Where we have used P = UI → W = V A . Similarly we can find with R = U/I → Ω = V/A " r L 1 C 1 # = r V 2 A 2 = Ω (2) Now for ² we know u E = 0 . 5 ² E 2 and thus [ ² ] = J/ ( mV 2 ) using F = qE → [ E ] = N/C = V/m , where the latter equality stems from UI = P → Nm/s = V As = V C/s . Similarly we have u M = 1 / (2 μ ) B 2 → [ μ ] = V 2 s 2 / ( Jm ) using from the Lorentz force law F = q ( ~ E + ~v × ~ B ) → [ B ] = s/m [ E ] = V s/m 2 . Together 1 √ ² μ = r mV 2 J Jm V 2 s 2 = m s (3) In an analogous manner r μ ² = r mV 2 J V 2 s 2 Jm = r V 4 s 2 V 2 A 2 s 2 = Ω . (4) You can also derive this latter units from [ ² ] = [ C ] /m = F/m and [ μ ] = [ L ] /m = H/m . 2. (a) Differentiating the first equation with respect to x yields ∂ 2 v ( x, t ) / ∂x 2 = L 1 ∂ ∂t ∂i ( x, t ) ∂x R 1 ∂i ( x, t ) ∂x (5) It is important to note that we were allowed to permute the order of the two derivations in the second term. This is possible because we assume that all first and second derivatives are smooth functions of x, t as well as i ( x, t ) is 1 Now inserting the second equation for ∂i/∂x yields the telegrapher’s equation. ∂ 2 v ( x, t ) ∂x 2 = L 1 C 1 ∂ 2 v ( x, t ) ∂t 2 + ( R 1 C 1 + G 1 L 1 ) ∂v ( x, t ) ∂t + R 1 G 1 v ( x, t ) (6) (b) To find a wave equation for i ( x, t ) we use the same method, but differentiate the second equation first, replacing the ∂v ( x, t ) /∂x...
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 Fall '04
 POLLACK
 Capacitance, Magnetism, Energy, 1%

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