TS EAMCET 20-July-2022 Shift 2 Solved Paper

Given, frequency of electric field of EM wave,
f=30‌MHz=3×107‌Hz
Amplitude, E0=150V∕m
∴ Amplitude of magnetic field vector,
B0=‌

E0

c

=‌

150

3×108

=5×107T
Angular frequency,
ω‌=2πf=22π×3×107
‌=6π×107‌rad∕ s
∴ Propagation constant,
k=‌

2π

λ

=‌

2π

c f

=‌

2πf

c

=‌

ω

c

=‌

6π×107

3×108

=‌

π

5

Since, E, B and direction of propagation are mutually perpendicular to each other, hence, for the given situation B will be directed along Z-axis ∴ Expression of magnetic field,
‌B=B0sin‌[kx−ωt]‌

^

z

‌⇒‌‌B=5×10−7sin‌[‌

π

5

x−6π×107t]‌hat‌z
‌=5×10−7sin‌[π(‌

x

5

−6×107t)]‌hat‌z
‌