Chapter 31: Q. 46 (page 903)

The electric field of a $450 MHz$ radio wave has a maximum rate of change of $4.5 \times 10^{11} (V / m) / s$ . What is the wave's magnetic field amplitude $?$

Short Answer

Expert verified

The waves magnetic field amplitude is $0.523 μT$ .

Step by step solution

Expression of electric field strength

Electric field strength is,

$E_{y} = E_{o} \sin (2 π (\frac{x}{λ} - ft))$

$\frac{{dE}_{y}}{dt} = - 2 {πfE}_{o} \cos (2 π (\frac{x}{λ} - ft))$

$E_{o} = \frac{(\frac{{dE}_{y}}{dt})}{2 πf}$

Calculation of magnetic field amplitude

Given info:

Maximum rate of change is $4.5 \times 10^{11} (V / m) / s$ .

Frequency is $450 MHz$ .

Magnetic field amplitude is,

$B_{o} = \frac{E_{o}}{c}$

Substitute electric field strength expression,

We get,

$B_{o} = \frac{(\frac{{dE}_{y}}{dt})}{2 πfc}$

Substitute all values,

$B_{o} = \frac{4.5 \times 10^{11} (V / m) / s}{2 π (450 \times 10^{6} Hz) (3 \times 10^{8} m / s)}$

$B_{o} = 0.523 μT$

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The electric field of a $450 MHz$ radio wave has a maximum rate of change of $4.5 \times 10^{11} (V / m) / s$ . What is the wave's magnetic field amplitude $?$

Short Answer

Step by step solution

Expression of electric field strength

Calculation of magnetic field amplitude

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The electric field of a 450MHz radio wave has a maximum rate of change of4.5×1011(V/m)/s . What is the wave's magnetic field amplitude?

Short Answer

Step by step solution

Expression of electric field strength

Calculation of magnetic field amplitude

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Solid State Physics

Magnetism

Engineering Physics

Mechanics and Materials

Classical Mechanics

Electricity

Study anywhere. Anytime. Across all devices.

The electric field of a $450 MHz$ radio wave has a maximum rate of change of $4.5 \times 10^{11} (V / m) / s$ . What is the wave's magnetic field amplitude $?$