Magnetic field in a plane electromagnetic wave is given by
` vecB = B_0 sin ( kx + omega t ) hatj T`
expression for corresponding electric field will be where c is speed of light .

Options

A. `vecE = B_0 c sin ( kx + omega t) hatk V//m`
B. `vecE = (B_0)/(c ) sin ( kx + omega t) V//m`
C. `vecE =- B_0 C sin ( kx + omega t) hatk V//m`
D. `vecE = B_0 C sin ( kx + omega t) hatk V//m`

Explanation

To find the expression for the electric field corresponding to the given magnetic field in a plane electromagnetic wave, we can follow these steps:

### Step 1: Identify the Given Magnetic Field
The magnetic field is given as:
\[ \vec{B} = B_0 \sin(kx + \omega t) \hat{j} \]
This indicates that the magnetic field oscillates in the y-direction (as represented by \(\hat{j}\)).

### Step 2: Determine the Direction of Wave Propagation
In electromagnetic waves, the direction of propagation is given by the wave vector \( \vec{k} \). Since the magnetic field varies with \(x\), the wave is propagating in the x-direction. Thus:
\[ \vec{k} \text{ is in the } \hat{i} \text{ direction.} \]

### Step 3: Use the Relationship Between Electric and Magnetic Fields
The relationship between the magnitudes of the electric field \(E_0\) and the magnetic field \(B_0\) in an electromagnetic wave is given by:
\[ \frac{E_0}{B_0} = c \]
where \(c\) is the speed of light. Therefore, we can express \(E_0\) as:
\[ E_0 = B_0 c \]

### Step 4: Determine the Direction of the Electric Field
The electric field \(\vec{E}\), the magnetic field \(\vec{B}\), and the direction of wave propagation \(\vec{k}\) are mutually perpendicular. To find the direction of \(\vec{E}\), we can use the right-hand rule. Since \(\vec{B}\) is in the \(\hat{j}\) direction and \(\vec{k}\) is in the \(\hat{i}\) direction, the electric field \(\vec{E}\) will be in the \(\hat{k}\) direction (out of the plane formed by \(\hat{i}\) and \(\hat{j}\)).

### Step 5: Write the Expression for the Electric Field
Now we can write the expression for the electric field:
\[ \vec{E} = E_0 \sin(kx + \omega t) \hat{k} \]
Substituting \(E_0\) from Step 3:
\[ \vec{E} = (B_0 c) \sin(kx + \omega t) \hat{k} \]

### Final Expression
Thus, the expression for the corresponding electric field is:
\[ \vec{E} = B_0 c \sin(kx + \omega t) \hat{k} \]