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NEET PHYSICSELECTROMAGNETIC WAVESEasy

Question

In a plane electromagnetic wave travelling in free space, the electric field component oscillates sinusoidally at a frequency of 2.0×1010 Hz2.0 \times 10^{10} \text{ Hz} and amplitude 48 V m148 \text{ V m}^{-1}. Then the amplitude of the oscillating magnetic field is: (Speed of light in free space =3×108 m s1= 3 \times 10^8 \text{ m s}^{-1})

A

1.6×106 T1.6 \times 10^{-6} \text{ T}

B

1.6×109 T1.6 \times 10^{-9} \text{ T}

C

1.6×108 T1.6 \times 10^{-8} \text{ T}

D

1.6×107 T1.6 \times 10^{-7} \text{ T}

Step-by-Step Solution

  1. Formula: For an electromagnetic wave propagating in free space, the amplitude of the magnetic field (B0B_0) is related to the amplitude of the electric field (E0E_0) and the speed of light (cc) by the relationship: B0=E0cB_0 = \frac{E_0}{c} (Reference: NCERT Physics Class 12, Chapter 8, Electromagnetic Waves).
  2. Given Data:
  • Electric Field Amplitude, E0=48 V m1E_0 = 48 \text{ V m}^{-1}.
  • Speed of light, c=3×108 m s1c = 3 \times 10^8 \text{ m s}^{-1}.
  • Frequency, ν=2.0×1010 Hz\nu = 2.0 \times 10^{10} \text{ Hz} (Note: Frequency is not required to find the amplitude ratio).
  1. Calculation: Substituting the values into the formula: B0=483×108 TB_0 = \frac{48}{3 \times 10^8} \text{ T} B0=16×108 TB_0 = 16 \times 10^{-8} \text{ T} Expressing in standard scientific notation: B0=1.6×107 TB_0 = 1.6 \times 10^{-7} \text{ T}

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ELECTROMAGNETIC WAVES. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSELECTROMAGNETIC WAVESelectromagnetictravellingelectriccomponentoscillates

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