## College Physics (4th Edition)

Published by McGraw-Hill Education

# Chapter 22 - Problems - Page 864: 27

#### Answer

(a) The amplitude of the electric field is $7.5~mV/m$ The frequency of the electric field is $3.0~MHz$ (b) The magnitude of the electric field is $4.5~mV/m$ The electric field is directed in the +x-direction.

#### Work Step by Step

(a) We can find the amplitude of the electric field: $E = c~B$ $E = (3.0\times 10^8~m/s)~(2.5\times 10^{-11}~T)$ $E = 7.5~mV/m$ The amplitude of the electric field is $7.5~mV/m$ Since the frequency of the electric field is the same as the frequency of the magnetic field, the frequency of the electric field is $3.0~MHz$ (b) We can find the magnitude of the electric field: $E = c~B$ $E = (3.0\times 10^8~m/s)~(1.5\times 10^{-11}~T)$ $E = 4.5~mV/m$ The magnitude of the electric field is $4.5~mV/m$ The direction of motion of the EM wave is determined by the cross-product $E\times B$. If the magnetic field points in the +z-direction, and the wave is traveling in the -y-direction, then the electric field must be pointing in the +x-direction. Using the right-hand rule, note that $(+x) \times (+z) = -y$

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