HomeLearn Fast select Course
Show Unit/ChapterChange UnitChange SubjectLast Menu
NEET-XII-Physics

40: Electromagnetic Waves

with Solutions - page 2
Qstn# ii-5 Prvs-QstnNext-Qstn
  • #5
    An electromagnetic wave going through vacuum is described by
    E = E0 sin (kx - ωt); B = B0 sin (kx - ωt).
    Then,
    (a) E0 k = B0 ω
    (b) E0 B0 = ωk
    (c) E0 ω = B0 k
    (d) None of these
    digAnsr:   a
    Ans : (a) E0 k = B0 ω
    The relation between E0 and B0 is given by `` \frac{{E}_{0}}{{B}_{0}}=c`` ...(1)
    Here, c = speed of the electromagnetic wave
    The relation between ​ω (the angular frequency) and k (wave number):
    `` \frac{\omega }{k}=c`` ...(2)
    Therefore, from (1) and (2), we get:
    `` \frac{{E}_{0}}{{B}_{0}}=```` \frac{\omega }{k}=c``
    `` \Rightarrow {E}_{0}k={B}_{0}\omega ``
    Page No 339: