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

To find the amplitude of the magnetic field associated with a laser beam, given its intensity, we can use the relation between intensity and the amplitudes of the electric and magnetic fields. The intensity \( I \) of an electromagnetic wave is given by the formula:

\(I = \frac{1}{2} c \varepsilon_0 E_0^2\) 

where:

• \(I\) is the intensity of the wave.
• \(c = 3 \times 10^8 \ \text{m/s}\) is the speed of light in vacuum.
• \(\varepsilon_0 = 8.85 \times 10^{-12} \ \text{C}^2/\text{N m}^2\) is the permittivity of free space.
• \(E_0\) is the amplitude of the electric field.

We are required to find the amplitude of the magnetic field \(B_0\), which is related to \(E_0\) by:

\(B_0 = \frac{E_0}{c}\)

First, we solve for the electric field amplitude \(E_0\) using the intensity formula:

\(E_0^2 = \frac{2I}{c \varepsilon_0}\)

Substitute the given values:

\(E_0^2 = \frac{2 \times 4.0 \times 10^{14}}{3 \times 10^8 \times 8.85 \times 10^{-12}}\)

\(E_0^2 = \frac{8.0 \times 10^{14}}{2.655 \times 10^{-3}}\)

\(E_0^2 = 3.01 \times 10^{17}\)

Taking the square root to find \(E_0\):

\(E_0 = \sqrt{3.01 \times 10^{17}} \approx 5.49 \times 10^8 \ \text{V/m}\)

Now, using \(B_0 = \frac{E_0}{c}\):

\(B_0 = \frac{5.49 \times 10^8}{3 \times 10^8}\)

\(B_0 \approx 1.83 \ \text{T}\)

Based on the calculations, the amplitude of the magnetic field associated with the beam is approximately 1.83 T, corresponding to the option

$1.83$

.