For a beam emerging from a filter facing a floodlight, the magnetic field is given by, `B= B_0 sin(kz – wt)`, where, `B_0 = 1.2 xx 10^(-8) T, k = 1.2 . xx 10^7 m^(-1),omega = 3.6 xx 10^(15) s^(-1)`
Now, find the average intensity of the beam.

Explanation

To find the average intensity of the beam given the magnetic field \( B = B_0 \sin(kz - \omega t) \), we will use the formula for average intensity in terms of the peak magnetic field \( B_0 \):

### Step-by-Step Solution:

1. **Identify Given Values**:
- \( B_0 = 1.2 \times 10^{-8} \, T \)
- \( k = 1.2 \times 10^{7} \, m^{-1} \)
- \( \omega = 3.6 \times 10^{15} \, s^{-1} \)

2. **Formula for Average Intensity**:
The average intensity \( I_{\text{avg}} \) of an electromagnetic wave can be calculated using the formula:
\[ I_{\text{avg}} = \frac{1}{2} \frac{B_0^2 c}{\mu_0} \]
where:
- \( c \) is the speed of light in vacuum (\( c \approx 3 \times 10^8 \, m/s \))
- \( \mu_0 \) is the permeability of free space (\( \mu_0 \approx 4\pi \times 10^{-7} \, T \cdot m/A \))

3. **Substituting Values**:
First, we calculate \( B_0^2 \):
\[ B_0^2 = (1.2 \times 10^{-8})^2 = 1.44 \times 10^{-16} \, T^2 \]

Now, substituting into the intensity formula:
\[ I_{\text{avg}} = \frac{1}{2} \cdot \frac{1.44 \times 10^{-16} \cdot 3 \times 10^8}{4\pi \times 10^{-7}} \]

4. **Calculating the Denominator**:
Calculate \( 4\pi \):
\[ 4\pi \approx 12.566 \quad \text{(approximately)} \]
Thus,
\[ 4\pi \times 10^{-7} \approx 12.566 \times 10^{-7} \, T \cdot m/A \]

5. **Final Calculation**:
Now, substituting the values:
\[ I_{\text{avg}} = \frac{1.44 \times 10^{-16} \cdot 3 \times 10^8}{2 \cdot 12.566 \times 10^{-7}} \]
\[ = \frac{4.32 \times 10^{-8}}{25.132 \times 10^{-7}} \]
\[ = \frac{4.32}{25.132} \times 10^{-1} \, W/m^2 \]
\[ \approx 0.0172 \, W/m^2 \]

### Conclusion:
The average intensity of the beam is approximately:
\[ I_{\text{avg}} \approx 0.0172 \, W/m^2 \]