The magnetic field of a beam emerging from a filter facing a floodlight is given by `B_(0)=12xx 10^(-8) sin (1.20 xx10^(7) z-3.60xx10^(15) t`. What is the average intensity (in `W/m^2` ) of the beam? (Take `pi=3` )

Answer

To find the average intensity of the beam given the magnetic field \( B_0 = 12 \times 10^{-8} \sin(1.20 \times 10^{7} z - 3.60 \times 10^{15} t) \), we can follow these steps:

### Step 1: Identify the amplitude of the magnetic field
The amplitude \( B_0 \) is given as:
\[ B_0 = 12 \times 10^{-8} \, \text{T} \]

### Step 2: Use the formula for average intensity
The average intensity \( I \) of an electromagnetic wave can be calculated using the formula:
\[ I = \frac{B_0^2}{2 \mu_0 c} \]
where:
- \( \mu_0 \) is the permeability of free space, given by \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)
- \( c \) is the speed of light in vacuum, approximately \( c = 3 \times 10^{8} \, \text{m/s} \)

### Step 3: Substitute the values into the formula
First, calculate \( B_0^2 \):
\[ B_0^2 = (12 \times 10^{-8})^2 = 144 \times 10^{-16} = 1.44 \times 10^{-14} \, \text{T}^2 \]

Next, substitute \( B_0^2 \), \( \mu_0 \), and \( c \) into the intensity formula:
\[ I = \frac{1.44 \times 10^{-14}}{2 \times (4\pi \times 10^{-7}) \times (3 \times 10^{8})} \]

### Step 4: Calculate the denominator
First, calculate \( 2 \mu_0 c \):
\[ 2 \mu_0 c = 2 \times (4\pi \times 10^{-7}) \times (3 \times 10^{8}) = 24\pi \times 10^{1} \]
Using \( \pi \approx 3 \):
\[ 2 \mu_0 c \approx 24 \times 3 \times 10^{1} = 72 \times 10^{1} = 7.2 \times 10^{2} \]

### Step 5: Calculate the average intensity
Now substitute back into the intensity formula:
\[ I = \frac{1.44 \times 10^{-14}}{7.2 \times 10^{2}} = 2 \times 10^{-17} \, \text{W/m}^2 \]

### Final Answer
Thus, the average intensity of the beam is:
\[ I \approx 2 \times 10^{-17} \, \text{W/m}^2 \]