Question:easy

Amplitude of intensity of magnetic field of an electromagnetic wave in vacuum is 500 nT. What will be the amplitude of the intensity of electric field of the wave?

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In vacuum the field amplitudes obey \(E_0 = cB_0\); convert nanotesla to tesla first.
Updated On: Jul 10, 2026
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Solution and Explanation

Step 1: Use the field ratio.
For any point in a plane electromagnetic wave in free space, the ratio of the electric field magnitude to the magnetic field magnitude is constant and equals the speed of light: \(\dfrac{E_0}{B_0} = c\).

Step 2: Rearrange for the electric field amplitude.
\(E_0 = c\,B_0\).

Step 3: Convert the magnetic amplitude to tesla.
\(B_0 = 500\ \text{nT} = 5 \times 10^{-7}\ \text{T}\) (since \(1\ \text{nT} = 10^{-9}\ \text{T}\)).

Step 4: Multiply.
\(E_0 = (3 \times 10^{8}\ \text{m/s})(5 \times 10^{-7}\ \text{T}) = 150\ \text{V/m}\).

So the electric field oscillates with a peak value of 150 volts per metre.

\[\boxed{E_0 = 150\ \text{V/m}}\]
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