JEE MAIN - Physics (2023 - 10th April Evening Shift - No. 10)
The amplitude of magnetic field in an electromagnetic wave propagating along y-axis is $$6.0 \times 10^{-7} \mathrm{~T}$$. The maximum value of electric field in the electromagnetic wave is
$$6.0 \times 10^{-7} ~\mathrm{Vm}^{-1}$$
$$5 \times 10^{14} ~\mathrm{Vm}^{-1}$$
$$180 ~\mathrm{Vm}^{-1}$$
$$2 \times 10^{15} ~\mathrm{Vm}^{-1}$$
Explanation
In an electromagnetic wave, the maximum value of the electric field E is related to the maximum value of the magnetic field B by the equation
$E = cB$
where c is the speed of light in a vacuum, which is approximately $3 × 10^8 m/s$.
Given that the amplitude of the magnetic field B is $6.0 × 10^{-7} T$, we can substitute these values into the equation to find the maximum value of the electric field:
$E = (3 × 10^8 m/s) \times (6.0 × 10^{-7} T) = 180 V/m$
Therefore, 180 V/m is the correct answer.
$E = cB$
where c is the speed of light in a vacuum, which is approximately $3 × 10^8 m/s$.
Given that the amplitude of the magnetic field B is $6.0 × 10^{-7} T$, we can substitute these values into the equation to find the maximum value of the electric field:
$E = (3 × 10^8 m/s) \times (6.0 × 10^{-7} T) = 180 V/m$
Therefore, 180 V/m is the correct answer.
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