QJEEM23Ph182b: JEE Main 2023 Physics - Class 12 JEEM23Ph182b Solution: A plane electromagnetic wave of frequency propagates in free…

In free space, the relationship between the electric field (E) and the magnetic field (B) in an electromagnetic wave is given by:

$B = \frac{E}{c}$

where c is the speed of light in a vacuum, approximately equal to $3 \times 10^8 \mathrm{~m} / \mathrm{s}$ . We are given that the electric field is $\overrightarrow{\mathrm{E}}=6.6 \hat{j} \mathrm{~V} / \mathrm{m}$ . To find the magnetic field, we can first calculate the magnitude of B:

$B = \frac{6.6 \mathrm{~V} / \mathrm{m}}{3 \times 10^8 \mathrm{~m} / \mathrm{s}} = 2.2 \times 10^{-8} \mathrm{~T}$

Now, we need to find the direction of the magnetic field. Since the electromagnetic wave propagates in the x-direction, and the electric field is in the y-direction (j), the magnetic field should be in the z-direction (k) to satisfy the right-hand rule for electromagnetic waves. The direction of the magnetic field will be positive k (counter-clockwise rotation from x to y).

So, $\overrightarrow{\mathrm{B}} = 2.2 \times 10^{-8} \hat{k} \mathrm{T}$ .