JEE Main 18 March 2021 Shift 1 Solved Paper

Given, ‌‌f=100MHz
B=(2.0×10−8)

∧

k

T
The speed of the light in free space,
c=(3×10s)

∧

i

m∕s
(the direction of wave propagates in x-direction)
Magnitude of electric field, E=Bc=2×10−8×3×108
⇒ E=6V/m
If

∧

n

be the direction of E, then

∧

i

=

∧

n

×

∧

k

⇒‌‌

∧

n

=

∧

j

Hence, the electric field at this point is 6

∧

j

V/m.