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

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: √ 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 ...
View Full Document

This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

Ask a homework question - tutors are online