If `B = 3sin (2pi/T* (y + ct)hati)` then find out amplitude of electric field

Answer

To find the amplitude of the electric field from the given magnetic field \( B = 3 \sin \left( \frac{2\pi}{T} (y + ct) \right) \hat{i} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Magnetic Field:**
The magnetic field is given as:
\[ B = 3 \sin \left( \frac{2\pi}{T} (y + ct) \right) \hat{i} \]
Here, the amplitude of the magnetic field \( B_0 \) is \( 3 \, \text{T} \).

2. **Use the Relationship Between Electric and Magnetic Fields:**
In an electromagnetic wave, the relationship between the electric field \( E \) and the magnetic field \( B \) is given by:
\[ E = cB \]
where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \, \text{m/s} \).

3. **Calculate the Amplitude of the Electric Field:**
Substituting the amplitude of the magnetic field into the equation:
\[ E_0 = cB_0 \]
\[ E_0 = (3 \times 10^8 \, \text{m/s}) \times (3 \, \text{T}) \]
\[ E_0 = 9 \times 10^8 \, \text{V/m} \]

4. **Final Answer:**
The amplitude of the electric field \( E_0 \) is:
\[ E_0 = 9 \times 10^8 \, \text{V/m} \]