Question

For a plane electromagnetic wave travelling in free space along X-axis, the magnetic field at a particular point in space is $\vec{B} = 2.1 \times 10^{-8} \hat{k} \, T$. The magnitude of the electric field at this point is

• A. $0.7 Vm^{-1}$
• B. $18.9 Vm^{-1}$
• C. $1.7 Vm^{-1}$
• D. $6.3 Vm^{-1}$

Hint:

Remember: $E = cB$. Electric field is much stronger in magnitude (in SI units) than the magnetic field for an EM wave.

Answer 1

The Correct Option is D

Step 1: Formula connecting E and B:
For an electromagnetic wave in free space, the magnitudes of the electric field ($E$) and magnetic field ($B$) are related by the speed of light ($c$): \[ E = c B \] where $c = 3 \times 10^8 \, m/s$.
Step 2: Calculation:
Given $B = 2.1 \times 10^{-8} \, T$. \[ E = (3 \times 10^8) \times (2.1 \times 10^{-8}) \] \[ E = 3 \times 2.1 \times 10^{8-8} \] \[ E = 6.3 \times 1 = 6.3 \, Vm^{-1} \]
Step 3: Final Answer:
The magnitude of the electric field is $6.3 \, Vm^{-1}$.