For a plane electromagnetic wave, the magnetic field at a point x and time t is \overrightarrow{{\B}}({x}, {t})=[1.2 \times10^{-7} \sin (0.5 \times10^{3} {x}+1.5 \times10^{11} {t}) {{k^{\^}}}] {T} The instantaneous electric field \overrightarrow{{\E}} corresponding to {{\B^{\to}}} is : (speed of light {c}=3 \times10^{8} {\ms}^{-1} ) [6 Sep 2020 Shift 2]

Correct Answer is : →E(x,t)=[−36sin(0.5×103x+1.5×1011t)^j]vm

{{\E^{\to}}}({x}, {t})=[36 \sin (0.5 \times10^{3} {x}+1.5 \times10^{11} {t}) \stackrel{\^}{{k}}] {{v}}/{{m}}

{{\E^{\to}}}({x}, {t})=[-36 \sin (0.5 \times10^{3} {x}+1.5 \times10^{11} {t}) \stackrel{\^}{{j}}] {{v}}/{{m}}

\overrightarrow{{\E}}({x}, {t})=[36 \sin (1 \times10^{3} {x}+0.5 \times10^{11} {t}) \stackrel{\^}{{j}}] {{v}}/{{m}}

\overrightarrow{{\E}}({x}, {t})=[36 \sin (1 \times10^{3} {x}+1.5 \times10^{11} {t}) \stackrel{\^}{{j}}] {{v}}/{{m}}

JEE Mains 06-Sep-2020 Shift 2 Solved Paper

Section: Physics

Question : 3 of 75

Marks: +1, -0

For a plane electromagnetic wave, the magnetic field at a point x and time t is

→

(x,t)=[1.2×10−7‌sin(0.5×103x+1.5×1011t)‌

The instantaneous electric field

→

corresponding to

→

is :
(speed of light c=3×108ms−1 )

[6 Sep 2020 Shift 2]

Go to Question:

Prev Question

Next Question