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3 years ago

8

n electromagnetic wave in vacuum has an electric field amplitude of 611 V/m. Calculate the amplitude of the corresponding magnet

ic field.

1 answer:

enot [183]3 years ago

6 0

Answer:

The corresponding magnetic field is

Explanation:

From the question we are told that

The electric field amplitude is $E_o = 611\ V/m$

Generally the magnetic field amplitude is mathematically represented as

$B_o = \frac{E_o }{c }$

Where c is the speed of light with a constant value

$c = 3.0 *0^{8} \ m/s$

So

$B_o = \frac{611 }{3.0*10^{8}}$

$B_o = 2.0 4 *10^{-6} \ Vm^{-2} s$

Since 1 T is equivalent to $V m^{-2} \cdot s$

$B_o = 2.0 4 *10^{-6} \ T$

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Answer:

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Given:

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Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

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Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

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