An electromagnetic wave is moving along negative $z$ ($-z$) direction and at any instant of time, at a point,

Question

An electromagnetic wave is moving along negative $z$ ($-z$) direction and at any instant of time, at a point, its electric field vector is $3\hat j 	ext{ V/m}$. The corresponding magnetic field at that point and instant will be: (Take $c=3	imes10^8 	ext{ ms}^{-1}$)

Accepted Answer

1. **Magnitude Calculation:** The magnitude of the magnetic field $B$ is related to the electric field $E$ and the speed of light $c$ by the relation $B = E/c$.
   Given $E = 3 	ext{ V/m}$ and $c = 3 	imes 10^8 	ext{ m/s}$:
   $$B = \frac{3}{3 	imes 10^8} = 10^{-8} 	ext{ T}$$
   Converting to nanoTesla ($1 	ext{ nT} = 10^{-9} 	ext{ T}$):
   $$B = 10 	imes 10^{-9} 	ext{ T} = 10 	ext{ nT}$$
2. **Direction Calculation:** The direction of propagation of an electromagnetic wave is given by the vector cross product of the electric field direction and the magnetic field direction: $\hat{v} = \hat{E} 	imes \hat{B}$.
   - Given direction of propagation $\hat{v} = -\hat{k}$ (negative z-direction).
   - Given direction of electric field $\hat{E} = \hat{j}$.
   - We need to find $\hat{B}$ such that $\hat{j} 	imes \hat{B} = -\hat{k}$.
   - Using the cyclic properties of unit vectors ($\hat{i} 	imes \hat{j} = \hat{k}$, $\hat{j} 	imes \hat{k} = \hat{i}$, $\hat{k} 	imes \hat{i} = \hat{j}$), we know that $\hat{j} 	imes \hat{i} = -\hat{k}$.
   - Therefore, the direction of the magnetic field must be $\hat{i}$.
3. **Conclusion:** The magnetic field vector is $10\hat{i} 	ext{ nT}$.

Step-by-Step Solution

Practice All PHYSICS Questions