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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. 345. Since the direction of propagation, E , and B are mutually
perpendicular, we infer that the only nonzero 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 §175). 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. 3433 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.
 Fall '08
 SPRUNGER
 Physics

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