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Physics Question 22 – NEET-UG 2023

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In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0×1010 Hz and amplitude 48 V m−1. Then the amplitude of oscillating magnetic field is : (Speed of light in free space =3×108 m s−1)

Electromagnetic waves consist of oscillating electric and magnetic fields that are perpendicular to each other and to the direction of wave propagation.

Step 1: Recall the relationship between electric and magnetic field amplitudes✦ Active

In an electromagnetic wave, the amplitude of the electric field (E0), the amplitude of the magnetic field (B0), and the speed of light (c) are related by the equation c=E0B0.

💡 Teacher's Secret Hint

The frequency of the wave is given but is not required for this specific calculation.

Step 2: Rearrange the formula to solve for the magnetic field amplitude○ Expand

From the relationship c=E0B0, we can express the magnetic field amplitude as B0=E0c.

Step 3: Substitute the given values and calculate○ Expand

Given E0=48 V m−1 and c=3×108 m s−1.

B0=48 V m−13×108 m s−1 B0=16×10−8 T B0=1.6×10−7 T
💡 Teacher's Secret Hint

Ensure units are consistent (SI units) for the calculation.

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