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Unformatted text preview: Physics 204 Homework 4 2. The current is I = 16 C / 1 . 5 × 10 3 s = 1 . 1 × 10 4 A. Plugging this and the distance r = 20 m into the formula for the magnetic field of a long current, one obtains B = μ I 2 πr = 1 . 1 × 10 4 T . 3. (Ch 10, P 1) (a) Equating the magnitudes of the magnetic fields B 1 of I 1 and B 2 of I 2 gives μ I 1 2 πr 1 = μ I 2 2 πr 2 . Solving this for I 2 , one obtains I 2 = r 2 r 1 I 1 = 8 m 2 m · 12 A = 72 A . The direction of I 2 must be the same as that of I 1 in order to make the magnetic field zero at the point in question. (b) As one can see from the solution above, all what one needs to find I 2 is I 1 the ratio r 2 r 1 . If one were first to compute B 1 numerically, one has to go through unnecessary calculations. 4. Only the horizontal component of the magnetic field contributes to the magnetic force on the current, since the current is parallel to the vertical component. The magnetic force is F B = ILB sin θ = (31 A) · (25 m) · (1 . 8 × 10...
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 Spring '09
 rollino
 Physics

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