Q.40. A magnetic field vector in an electromagnetic wave is represented by, . Its associated electri

Q: What is the correct answer for JEE Main Physics Q40?

The correct answer is A. Sol. In an electromagnetic wave, the electric field $\vec{E}$, magnetic field $\vec{B}$, and direction of propagation are mutually perpendicular. The direction of propagation is given by the wave vector $\vec{k}$ which is in the positive x-direction (from $kx = \frac{2\pi x}{\lambda}$). The relation between $\vec{E}$, $\vec{B}$, and the direction of propagation is $\vec{E} \times \vec{B}$ is in th

Q: Q.40. A magnetic field vector in an electromagnetic wave is represented by, . Its associated electric field vector is

The correct answer is A. Sol. In an electromagnetic wave, the electric field $\vec{E}$, magnetic field $\vec{B}$, and direction of propagation are mutually perpendicular. The direction of propagation is given by the wave vector $\vec{k}$ which is in the positive x-direction (from $kx = \frac{2\pi x}{\lambda}$). The relation between $\vec{E}$, $\vec{B}$, and the direction of propagation is $\vec{E} \times \vec{B}$ is in th

Q: What are the options for this JEE Main Physics question?

A. $\vec{E} = -\nu \lambda B_0 \sin \left( 2\pi \nu t - \frac{2\pi x}{\lambda} \right)$ · B. $\vec{E} = -\nu \lambda B_0 \sin \left( 2\pi \nu t - \frac{2\pi x}{\lambda} \right)$ · C. $\vec{E} = \nu \lambda B_0 \sin \left( 2\pi \nu t - \frac{2\pi x}{\lambda} \right)$ · D. $\vec{E} = \nu \lambda B_0 \sin \left( 2\pi \nu t - \frac{2\pi x}{\lambda} \right)$

Question

Q.40. A magnetic field vector in an electromagnetic wave is represented by, . Its associated electric field vector is

FirstInTest · Accepted Answer

Sol. In an electromagnetic wave, the electric field $\vec{E}$, magnetic field $\vec{B}$, and direction of propagation are mutually perpendicular. The direction of propagation is given by the wave vector $\vec{k}$ which is in the positive x-direction (from $kx = \frac{2\pi x}{\lambda}$). The relation between $\vec{E}$, $\vec{B}$, and the direction of propagation is $\vec{E} 	imes \vec{B}$ is in the direction of propagation. Also, $E_0 = c B_0 = 
u \lambda B_0$. Given $\vec{B}$ is in the y-direction (from the options, assuming $B_0$ is a scalar magnitude and the vector is implied by the options). Let's assume $\vec{B} = B_0 \sin(2\pi 
u t - kx) \hat{j}$. The wave propagates in the +x direction. So $\vec{E}$ must be in the z-direction. $\hat{k} 	imes \hat{j} = -\hat{i}$. So $\vec{E}$ must be in the $-\hat{k}$ direction. $\vec{E} = -E_0 \sin(2\pi 
u t - kx) \hat{k}$. $E_0 = cB_0 = 
u \lambda B_0$. So $\vec{E} = -
u \lambda B_0 \sin \left( 2\pi 
u t - \frac{2\pi x}{\lambda} ight) \hat{k}$. This matches option (1) if we assume the vector direction is implied by the options. The solution states $\hat{E} 	imes \hat{B} = \hat{C}$ where $\hat{C}$ is the direction of propagation. If $\hat{B} = \hat{j}$ (y-direction) and propagation is $\hat{i}$ (x-direction), then $\hat{E}$ must be in the $-\hat{k}$ (negative z-direction) because $(-\hat{k}) 	imes \hat{j} = -(-\hat{i}) = \hat{i}$. So $\vec{E} = -
u \lambda B_0 \sin \left( 2\pi 
u t - \frac{2\pi x}{\lambda} ight)$. This matches option (1).

Answer

$\vec{E} = -
u \lambda B_0 \sin \left( 2\pi 
u t - \frac{2\pi x}{\lambda} ight)$

Answer

$\vec{E} = -
u \lambda B_0 \sin \left( 2\pi 
u t - \frac{2\pi x}{\lambda} ight)$

Answer

$\vec{E} = 
u \lambda B_0 \sin \left( 2\pi 
u t - \frac{2\pi x}{\lambda} ight)$

Answer

$\vec{E} = 
u \lambda B_0 \sin \left( 2\pi 
u t - \frac{2\pi x}{\lambda} ight)$

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