p34_071

# p34_071 - √ 71. (a) The electric ﬁeld amplitude is Em =...

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Unformatted text preview: √ 71. (a) The electric ﬁeld amplitude is Em = 2 Erms = 70.7 V/m, so that the magnetic ﬁeld amplitude is Bm = 2.36 × 10−7 T by Eq. 34-5. Since the direction of propagation, E , and B are mutually perpendicular, we infer that the only non-zero component of B is Bx , and note that the direction of propagation being along the −z axis means the spatial and temporal parts of the wave function argument are of like sign (see §17-5). Also, from λ = 250 nm, we ﬁnd that f = c/λ = 1.20 × 1015 Hz, which leads to ω = 2πf = 7.53 × 1015 rad/s. Also, we note that k = 2π/λ = 2.51 × 107 m−1 . Thus, assuming some “initial condition” (that, say the ﬁeld is zero, with its derivative positive, at z = 0 when t = 0), we have Bx = 2.36 × 10−7 sin 2.51 × 107 z + 7.53 × 1015 t in SI units. √ (b) The exposed area of the triangular chip is A = 3 2 /8, where = 2.00 × 10−6 m. The intensity of the wave is 1 2 I= E 2 = 6.64 W/m . cµ0 rms Thus, Eq. 34-33 leads to F= 2IA = 3.83 × 10−20 N . c ...
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## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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