An electromagnetic wave travelling in the x- direction has frequency of 2× 10^14Hz and electric field amplitude of 27Vm^–1 oscillates in Y-direction.…
JEE Main 2015 — Physics Electromagnetism
2015mcqmedium
An electromagnetic wave travelling in the x− direction has frequency of 2×1014Hz and electric field amplitude of 27Vm–1 oscillates in Y−direction. From the options given below, which one describes the magnetic field for this wave?
Official previous-year question
Held on 10 Apr 2015 · Verified 6 Jul 2026.
Options
A
B(x,t)=(9×10−8T)j^sin[1.5×10−6x−2×1014t]
B
B(x,t)=(9×10−8T)i^sin[2π(1.5×10−8x−2×1014t)]
C
B(x,t)=(3×10−8T)j^sin2π[(1.5×10−8x)−2×1014t]
D
B(x,t)=(9×10−8T)k^sin2π[(1.5×10−6x)−2×1014t]
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Solution
When undefined
Then B=B0sin(kx−ωt)
Of light in travelling along i^ then B in either along j^ or k^.
∴ speed of light C=B0E0⇒B0=CE0
⇒B0=3×10827=9×10−8T
also, ω=2πf=2π×2×1014=4π×1014
k=cω=3×1084π×1014
=(1.5×10−62π)
Looking into the option the correct
Answer is B=9×10−8sin2π(1.5×10−6x−2×1014t)k^
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