Verified Solution • 2025–26 Session moderate

QJEEM22Ph988d: JEE Main 2022 Physics - Class 12 JEEM22Ph988d Solution: Light wave travelling in air along x-direction is given by .…

Light wave travelling in air along x-direction is given by Ey=540sin⁡π×104(x−ct) Vm−1{E_y} = 540\sin \pi \times {10^4}(x - ct)\,V{m^{ - 1}} . Then, the peak value of magnetic field of wave will be (Given c = 3 ×\times 108 ms −- 1)

Multiple Choice Options

  • A 18 ×\times 10 −- 7 T
  • B 54 ×\times 10 −- 7 T
  • C 54 ×\times 10 −- 8 T
  • D 18 ×\times 10 −- 8 T

Step-by-Step Verified Solution

c=3×108c = 3 \times {10^8} m/sec

B=Ec=5403×108=18×10−7 TB = {E \over c} = {{540} \over {3 \times {{10}^8}}} = 18 \times {10^{ - 7}}\,T

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring