
The magnetic field in a travelling electromagnetic wave has a peak value of 20 nT. The peak value of electric field strength is
A: 6V/m
B: 9V/m
C: 12V/m
D: 3V/m
Answer
472.8k+ views
Hint: The peak value of electric field strength is the product of peak value of magnetic field and velocity of light. Thus by substituting the value of magnetic field and velocity of light we get the value of peak electric field strength. So in an EM wave, the magnitude of electric field is proportional to the magnitude of magnetic field.
Formula used:
${{E}_{0}}=c{{B}_{0}}$
where, ${{E}_{0}}$ is the electric field
${{B}_{0}}$ is the magnetic field
c is the velocity of light.
Complete answer:
By applying the equation we get,
${{E}_{0}}=c{{B}_{0}}$
where, ${{E}_{0}}$ is the electric field
${{B}_{0}}$ is the magnetic field
c is the velocity of light.
Substituting the values in the above equation,
${{E}_{0}}=3\times {{10}^{8}}\times 20\times {{10}^{-9}}=6V/m$
So, the correct answer is “Option A”.
Additional Information:
Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force. The electric field generally goes outward from a positive charge and goes in toward a negative point charge. The magnitude and direction of the electric field can be expressed by the value of E, called electric field strength or electric field intensity. Electric field lines always flow from a positive to the negative charges.
Note:
In an electromagnetic wave, the electric field and magnetic field carry the same energy. So in an EM wave, the magnitude of electric field is proportional to the magnitude of magnetic field. And the proportionality constant is called velocity of light. Thus the peak value of electric field strength is the product of peak value of magnetic field and velocity of light.
Formula used:
${{E}_{0}}=c{{B}_{0}}$
where, ${{E}_{0}}$ is the electric field
${{B}_{0}}$ is the magnetic field
c is the velocity of light.
Complete answer:
By applying the equation we get,
${{E}_{0}}=c{{B}_{0}}$
where, ${{E}_{0}}$ is the electric field
${{B}_{0}}$ is the magnetic field
c is the velocity of light.
Substituting the values in the above equation,
${{E}_{0}}=3\times {{10}^{8}}\times 20\times {{10}^{-9}}=6V/m$
So, the correct answer is “Option A”.
Additional Information:
Electric field is defined as the electric force per unit charge. The direction of the field is taken to be the direction of the force. The electric field generally goes outward from a positive charge and goes in toward a negative point charge. The magnitude and direction of the electric field can be expressed by the value of E, called electric field strength or electric field intensity. Electric field lines always flow from a positive to the negative charges.
Note:
In an electromagnetic wave, the electric field and magnetic field carry the same energy. So in an EM wave, the magnitude of electric field is proportional to the magnitude of magnetic field. And the proportionality constant is called velocity of light. Thus the peak value of electric field strength is the product of peak value of magnetic field and velocity of light.
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