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Chapter 15: Problem 2149

In an electromagnetic wave, if the amplitude of magnetic field is $3 \times 10^{-10} \mathrm{~T}$, the amplitude of the associated electric field will be (A) \(9 \times 10^{-2} \overline{\mathrm{Vm}^{-1}}\) (B) \(3 \times 10^{-10} \mathrm{Vm}^{-1}\) (C) \(3 \times 10^{-2} \mathrm{Vm}^{-1}\) (D) \(1 \times 10^{-18} \mathrm{Vm}^{-1}\)

Short Answer

Expert verified
The amplitude of the associated electric field is \(9 \times 10^{-2} \mathrm{Vm}^{-1}\) (Option A).

Step by step solution

01

Write down the given values

The amplitude of the magnetic field is given as \(B = 3 \times 10^{-10} T\).
02

Write down the relationship between electric and magnetic field amplitudes

The relationship between electric and magnetic field amplitudes is given by the formula: \(E = cB\), where \(E\) is the amplitude of the associated electric field, \(B\) is the amplitude of the magnetic field, and \(c\) is the speed of light in a vacuum (\(3\times 10^8 m/s\)).
03

Substitute the given values into the formula

Substitute the given value of the magnetic field amplitude \(B = 3 \times 10^{-10} T\) and the speed of light \(c = 3 \times 10^8 m/s\) into the formula: \(E = (3 \times 10^8 m/s)(3 \times 10^{-10} T)\)
04

Calculate the electric field amplitude

Multiply the values in the expression: \(E = 9 \times 10^{-2} Vm^{-1}\)
05

Compare the result with the given options

Comparing the result \(E = 9 \times 10^{-2} Vm^{-1}\) with the given options, we see that it matches option (A). The amplitude of the associated electric field is \(9 \times 10^{-2} Vm^{-1}\) (Option A).

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Most popular questions from this chapter

A plane electromagnetic wave is incident on a material surface. If the wave delivers momentum \(p\) and energy \(E\), then (A) \(p=0, E=0\) (B) \(p \neq 0, E \neq 0\) (C) \(p \neq 0, E=0\) (D) \(p=0, E \neq 0\)

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