Back to Directory
NEET PHYSICSMedium

The magnetic field of a plane electromagnetic wave is given by, B⃗=3×10−8cos⁡(1.6×103x+48×1010t)j^ T\vec B=3\times10^{-8}\cos(1.6\times10^3x+48\times10^{10}t)\hat j \text{ T}. The associated electric field will be:

A

3×10−8cos⁡(1.6×103x+48×1010t)i^ V/m3\times10^{-8}\cos(1.6\times10^3x+48\times10^{10}t)\hat i \text{ V/m}

B

3×10−8sin⁡(1.6×103x+48×1010t)i^ V/m3\times10^{-8}\sin(1.6\times10^3x+48\times10^{10}t)\hat i \text{ V/m}

C

9sin⁡(1.6×103x−48×1010t)k^ V/m9\sin(1.6\times10^3x-48\times10^{10}t)\hat k \text{ V/m}

D

9cos⁡(1.6×103x+48×1010t)k^ V/m9\cos(1.6\times10^3x+48\times10^{10}t)\hat k \text{ V/m}

Step-by-Step Solution

  1. Calculate Electric Field Amplitude (E0E_0): The amplitude of the electric field is related to the magnetic field amplitude (B0B_0) by the speed of light (cc): E0=cB0E_0 = c B_0. Given B0=3×10−8 TB_0 = 3 \times 10^{-8} \text{ T} and c=3×108 m/sc = 3 \times 10^8 \text{ m/s}: E0=(3×108)×(3×10−8)=9 V/mE_0 = (3 \times 10^8) \times (3 \times 10^{-8}) = 9 \text{ V/m}
  2. Determine Direction: The direction of propagation of an EM wave is given by E⃗×B⃗\vec E \times \vec B. The argument (kx+ωt)(kx + \omega t) indicates the wave is travelling in the negative x-direction (−i^-\hat i). The magnetic field is in the y-direction (j^\hat j). We need a direction for E⃗\vec E such that E^×j^=−i^\hat E \times \hat j = -\hat i. Using vector cross product rules, k^×j^=−i^\hat k \times \hat j = -\hat i. Thus, the electric field is along the z-axis (k^\hat k).
  3. Phase: In a plane electromagnetic wave in free space, the electric and magnetic fields oscillate in phase. Therefore, the function remains cosine with the same argument.
  4. Conclusion: Combining the magnitude, direction, and phase, E⃗=9cos⁡(1.6×103x+48×1010t)k^ V/m\vec E = 9 \cos(1.6\times10^3x+48\times10^{10}t)\hat k \text{ V/m}.
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started