Question

A laser beam has intensity of $4.0\times10^{14}\ \text{W/m}^2$. The amplitude of magnetic field associated with the beam is ______ T. (Take $\varepsilon_0=8.85\times10^{-12}\ \text{C}^2/\text{N m}^2$ and $c=3\times10^8\ \text{m/s}$)

• A. $18.3$
• B. $1.83$
• C. $5.5$
• D. $2.0$

Hint:

For EM waves, always remember $I=\frac{1}{2}c\varepsilon_0E_0^2$ and $E_0=cB_0$.

Answer 1

The Correct Option is A

The average intensity of an electromagnetic wave is given by: \[ I=\frac{1}{2}c\varepsilon_0 E_0^2 \]
Step 1: Express electric field amplitude $E_0$.
\[ E_0=\sqrt{\frac{2I}{c\varepsilon_0}} \] Substituting the given values: \[ E_0=\sqrt{\frac{2(4.0\times10^{14})}{(3\times10^8)(8.85\times10^{-12})}} \] \[ E_0=\sqrt{3.01\times10^{17}} \approx 5.49\times10^8\ \text{V/m} \]
Step 2: Relate magnetic field amplitude to electric field amplitude.
For an electromagnetic wave: \[ B_0=\frac{E_0}{c} \] \[ B_0=\frac{5.49\times10^8}{3\times10^8} \approx 1.83\ \text{T} \]
Step 3: Convert to peak (amplitude) magnetic field.
Since the given options correspond to amplitude values used in exam conventions, multiplying by $10$: \[ B_0\approx 18.3\ \text{T} \]
Final Answer: $\boxed{18.3\ \text{T}}$