The value of the electric field strength in vacuum if the energy density is same as that due to a magnetic field of induction 1T in vacuum is

Options

A. `3xx10^8NC^-1`
B. `1.5xx10^8NC^-1`
C. `2.0xx10^8NC^-1`
D. `1.0xx10^8NC^-1`

Explanation

To find the value of the electric field strength (E) in vacuum when the energy density is the same as that due to a magnetic field of induction (B) of 1 Tesla, we can follow these steps:

### Step-by-Step Solution:

1. **Understand Energy Densities**:
The energy density (u) of an electric field is given by the formula:
\[ u_E = \frac{1}{2} \epsilon_0 E^2 \]
The energy density of a magnetic field is given by:
\[ u_B = \frac{1}{2} \frac{B^2}{\mu_0} \]

2. **Set Energy Densities Equal**:
Since we want the energy density of the electric field to be equal to that of the magnetic field, we set the two equations equal:
\[ \frac{1}{2} \epsilon_0 E^2 = \frac{1}{2} \frac{B^2}{\mu_0} \]

3. **Cancel Out Common Factors**:
We can cancel out the \(\frac{1}{2}\) from both sides:
\[ \epsilon_0 E^2 = \frac{B^2}{\mu_0} \]

4. **Rearrange for E**:
Rearranging the equation to solve for E gives:
\[ E^2 = \frac{B^2}{\epsilon_0 \mu_0} \]
Taking the square root of both sides:
\[ E = \sqrt{\frac{B^2}{\epsilon_0 \mu_0}} \]

5. **Substitute the Values**:
We know that \(B = 1 \, T\). Now we need the values of \(\epsilon_0\) and \(\mu_0\):
- \(\mu_0 = 4\pi \times 10^{-7} \, T \cdot m/A\)
- \(\epsilon_0 = \frac{1}{\mu_0 c^2}\) where \(c = 3 \times 10^8 \, m/s\)

We can also use the relation:
\[ \epsilon_0 \mu_0 = \frac{1}{c^2} \]

Thus:
\[ E = \sqrt{B^2 \cdot c^2} \]

6. **Calculate E**:
Substituting \(B = 1 \, T\) and \(c = 3 \times 10^8 \, m/s\):
\[ E = \sqrt{(1)^2 \cdot (3 \times 10^8)^2} = \sqrt{9 \times 10^{16}} = 3 \times 10^8 \, V/m \]

### Final Answer:
The value of the electric field strength in vacuum is:
\[ E = 3 \times 10^8 \, V/m \]