QJEEM19Phf7e1: JEE Main 2019 Physics - Class 12 JEEM19Phf7e1 Solution: The magnetic field of an electromagnetic wave is given by :-…

Question

Verified Solution • 2025–26 Session moderate

The magnetic field of an electromagnetic wave is given by :-

\mathop B\limits^ \to = 1.6 \times {10^{ - 6}}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){{Wb} \over {{m^2}}}

The associated electric field will be :-

A $\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( -2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}$
B $\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z - 6 \times {{10}^{15}}t} \right)\left( 2{\mathop i\limits^ \wedge + \mathop {j}\limits^ \wedge } \right){V \over m}$
C $\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {\mathop i\limits^ \wedge - \mathop {2j}\limits^ \wedge } \right){V \over m}$
D $\mathop E\limits^ \to = 4.8 \times {10^2}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( -{\mathop i\limits^ \wedge + \mathop {2j}\limits^ \wedge } \right){V \over m}$

Accepted Answer

If we use that direction of light propagation will be along

\overrightarrow E \times \overrightarrow B

. Then (A) option is correct.
Magnitude of E = CB

E = 3 × 10⁸ × 1.6 × 10^–6 ×

\sqrt 5

E = 4.8 ×

{10^{2\sqrt 5 }}

\overrightarrow E

and

\overrightarrow B

are perpendicular to each other

\Rightarrow \overrightarrow E .\overrightarrow B = 0

\Rightarrow

Either direction of

\overrightarrow E

\widehat i - 2\widehat j

or

-\widehat i + 2\widehat j

from given option

Also wave propagation direction is parallel to

\overrightarrow E \times \overrightarrow B

which is

- \widehat k

\Rightarrow

\overrightarrow E

is along (

-\widehat i + 2\widehat j

)

Step-by-Step Verified Solution