JEE Main 16 March 2021 Shift 1 Solved Paper

Given, frequency, f=500MHz=5×108Hz
B=8.0×10−8

∧

Z

T
∴ Magnitude of peak value of magnetic field is given by
B0=8×10−8T
We know that, ‌

E0

B0

=C
where, E0 is the magnitude of peak value of electric field and c is the speed of electromagnetic wave in air (or vacuum).
⇒‌‌E0=cB0
=3×108×8×10−8=24V∕m
Since, the direction of propagation of electromagnetic wave is perpendicular to the direction of E and B both.
∴ Direction of propagation is given by

∧

E

×

∧

B

.
As, the wave is travelling in y-direction,
and the magnetic field is in z-direction.
⇒

∧

E

×

∧

z

‌‌=

∧

y

∧

E

‌‌=−

∧

x

∴ The value of electric field will be −24

∧

x

V∕m.