Answer:
[tex]E=6\times 10^5\ N/C[/tex]
Explanation:
It is given that,
Magnetic field, [tex]B=2\ mT=2\times 10^{-3}\ T[/tex]
Let E is the magnitude of electric field amplitude. We know that the relation between the magnitude of electric field and the magnetic filed is given by :
[tex]E=c\times B[/tex]
[tex]E=3\times 10^8\ m/s\times 2\times 10^{-3}\ T[/tex]
[tex]E=6\times 10^5\ N/C[/tex]
So, the electric field amplitude of an electromagnetic wave is [tex]6\times 10^5\ N/C[/tex]. Hence, this is the required solution.
The electric field amplitude of the electromagnetic wave with the given magnetic field amplitude is 6 × 10⁵ V/m.
Given the data in the question;
An electric field is field around particles that are electrically charged which exerts force on other particles which are also charged around it by attraction or repulsion.
Electric field amplitude is simply the maximum strength of the electric and magnetic fields.
It can be expressed as;
[tex]E_m = c * B_m[/tex]
Where c is the speed of light ( [tex]c = 3 * 10^8m/s[/tex] ) and [tex]B_m[/tex] is the magnetic field amplitude.
We substitute our values into the expression above
[tex]E_m = c * B_m\\\\E_m = 3*10^{8}m/s\ *\ 2*10^{-3}T\\\\E_m = 600,000V/m\\\\E_m = 6* 10^5V/m[/tex]
Therefore, the electric field amplitude of the electromagnetic wave with the given magnetic field amplitude is 6 × 10⁵ V/m.
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