In an electromagnetic wave, the average energy density associated with magnetic field is:

Options

A. `Li_(0)^(2)//2`
B. `B^(2)//2mu_(0)`
C. `mu_(0)B^(2)//2`
D. `mu_(0)//2B^(2)`

Answer

To find the average energy density associated with the magnetic field in an electromagnetic wave, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Total Energy Density**:
The total average energy density \( u \) in an electromagnetic wave is the sum of the average energy densities of the electric field \( u_E \) and the magnetic field \( u_B \):
\[ u = u_E + u_B \]

2. **Average Energy Density of Electric Field**:
The average energy density associated with the electric field \( u_E \) is given by the formula:
\[ u_E = \frac{1}{2} \epsilon_0 E^2 \]
where \( E \) is the electric field strength and \( \epsilon_0 \) is the permittivity of free space.

3. **Expressing in Terms of Amplitude**:
The average value of \( E^2 \) in terms of the amplitude \( E_0 \) (peak value) of the electric field is:
\[ \langle E^2 \rangle = \frac{E_0^2}{2} \]
Therefore, substituting this into the expression for \( u_E \):
\[ u_E = \frac{1}{2} \epsilon_0 \left(\frac{E_0^2}{2}\right) = \frac{1}{4} \epsilon_0 E_0^2 \]

4. **Average Energy Density of Magnetic Field**:
Similarly, the average energy density associated with the magnetic field \( u_B \) is given by:
\[ u_B = \frac{1}{2} \mu_0 B^2 \]
where \( B \) is the magnetic field strength and \( \mu_0 \) is the permeability of free space.

5. **Expressing in Terms of Amplitude**:
The average value of \( B^2 \) in terms of the amplitude \( B_0 \) (peak value) of the magnetic field is:
\[ \langle B^2 \rangle = \frac{B_0^2}{2} \]
Thus, substituting this into the expression for \( u_B \):
\[ u_B = \frac{1}{2} \mu_0 \left(\frac{B_0^2}{2}\right) = \frac{1}{4} \mu_0 B_0^2 \]

6. **Relating Electric and Magnetic Fields**:
In an electromagnetic wave, the electric field \( E \) and the magnetic field \( B \) are related by the equation:
\[ c = \frac{E}{B} \]
where \( c \) is the speed of light in vacuum. Therefore, we can express \( B \) in terms of \( E \):
\[ B = \frac{E}{c} \]

7. **Substituting to Find \( u_B \)**:
Substituting \( B = \frac{E}{c} \) into the expression for \( u_B \):
\[ u_B = \frac{1}{2} \mu_0 \left(\frac{E}{c}\right)^2 = \frac{1}{2} \mu_0 \left(\frac{E^2}{c^2}\right) \]

8. **Using the Relation \( c^2 = \frac{1}{\mu_0 \epsilon_0} \)**:
Since \( c^2 = \frac{1}{\mu_0 \epsilon_0} \), we can substitute this into the expression for \( u_B \):
\[ u_B = \frac{1}{2} \mu_0 \left(\frac{E^2}{\frac{1}{\mu_0 \epsilon_0}}\right) = \frac{1}{2} \epsilon_0 E^2 \]

9. **Final Expression for Average Energy Density of Magnetic Field**:
Therefore, the average energy density associated with the magnetic field can be expressed as:
\[ u_B = \frac{B^2}{2 \mu_0} \]

10. **Conclusion**:
Thus, the average energy density associated with the magnetic field in an electromagnetic wave is:
\[ \frac{B^2}{2 \mu_0} \]
Hence, the correct answer is option number 2.