If the electric field and magnetic field of an electromagnetic wave are related as `B= ( E)/( c)` where the symbols have their usual meanings and the energy in a given volume of space due to the electric field part is U, then the energy due to the magnetic field part will be

Options

A. `(U )/(c )`
B. `( U)/( c^2)`
C. `( U)/(2)`
D. U

Answer

To find the energy due to the magnetic field part of an electromagnetic wave, we can follow these steps:

### Step 1: Understand the relationship between electric and magnetic fields
We know that for an electromagnetic wave, the magnetic field \( B \) is related to the electric field \( E \) by the equation:
\[ B = \frac{E}{c} \]
where \( c \) is the speed of light.

### Step 2: Write the energy density formulas
The energy density \( U_e \) due to the electric field is given by:
\[ U_e = \frac{1}{2} \epsilon_0 E^2 \]
The energy density \( U_b \) due to the magnetic field is given by:
\[ U_b = \frac{1}{2} \frac{B^2}{\mu_0} \]

### Step 3: Substitute the expression for \( B \)
Using the relationship \( B = \frac{E}{c} \), we can substitute \( B \) into the formula for magnetic energy density:
\[ U_b = \frac{1}{2} \frac{\left(\frac{E}{c}\right)^2}{\mu_0} \]
This simplifies to:
\[ U_b = \frac{1}{2} \frac{E^2}{c^2 \mu_0} \]

### Step 4: Relate \( c^2 \) to \( \epsilon_0 \) and \( \mu_0 \)
We know that the speed of light \( c \) is given by:
\[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \implies c^2 = \frac{1}{\mu_0 \epsilon_0} \]
Thus, we can rewrite \( \frac{1}{c^2} \) as:
\[ \frac{1}{c^2} = \mu_0 \epsilon_0 \]

### Step 5: Substitute back into the magnetic energy density
Now substituting \( \frac{1}{c^2} \) into the equation for \( U_b \):
\[ U_b = \frac{1}{2} \frac{E^2}{\mu_0} \cdot \mu_0 \epsilon_0 = \frac{1}{2} \epsilon_0 E^2 \]

### Step 6: Relate \( U_b \) to \( U \)
From the problem statement, we know that the energy due to the electric field part is \( U \):
\[ U = \frac{1}{2} \epsilon_0 E^2 dV \]
Thus, we can conclude that:
\[ U_b = U \]

### Final Answer
The energy due to the magnetic field part will also be \( U \).

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