The magnetic field in the plane electromagnetic wave is given by
`B_(z)=2xx10^(-7) sin(0.5xx10^(3)x+1.5xx10^(11)t)` tesla.
The expression for electric field will be:

Options

A. `vecE=-60sin(0.5xx10^3x+1.5xx10^(10)t)hatjVm^(-1)`
B. `vecE=+60sin(0.5xx10^3x+1.5xx10^(10)t)hatjVm^(-1)`
C. `vecE=+60sin(0.5xx10^(11)x+1.5xx10^(3)t)hatjVm^(-1)`
D. `vecE=-60sin(0.5xx10^3x+1.5xx10^(11)t)hatjVm^(-1)`

Explanation

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

### Step 1: Identify the Given Magnetic Field
The magnetic field is given by:
\[ B_z = 2 \times 10^{-7} \sin(0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \text{ tesla} \]

### Step 2: Compare with General Form
The general form of the magnetic field in an electromagnetic wave is:
\[ B_z = B_0 \sin(kx + \omega t) \]
where \( B_0 \) is the amplitude, \( k \) is the wave number, and \( \omega \) is the angular frequency.

From the given equation, we can identify:
- \( B_0 = 2 \times 10^{-7} \) T
- \( k = 0.5 \times 10^{3} \) rad/m
- \( \omega = 1.5 \times 10^{11} \) rad/s

### Step 3: Determine the Direction of the Electric Field
In an electromagnetic wave, the electric field \( E \) is perpendicular to both the magnetic field \( B \) and the direction of wave propagation. Since the magnetic field is in the \( z \)-direction and the wave is propagating in the \( x \)-direction, the electric field will be in the \( y \)-direction.

### Step 4: Use the Relationship Between Electric and Magnetic Fields
The relationship between the electric field \( E \) and the magnetic field \( B \) in a vacuum is given by:
\[ E = cB \]
where \( c \) is the speed of light in vacuum, approximately \( 3 \times 10^8 \) m/s.

### Step 5: Calculate the Electric Field
Substituting the values we have:
\[ E = cB_0 \sin(kx + \omega t) \]
\[ E = (3 \times 10^8) \times (2 \times 10^{-7}) \sin(0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \]
\[ E = 6 \times 10^1 \sin(0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \text{ volts/m} \]
\[ E = 60 \sin(0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \text{ volts/m} \]

### Step 6: Specify the Direction
Since the electric field is in the \( y \)-direction, we can write:
\[ \vec{E} = 60 \hat{j} \sin(0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \text{ volts/m} \]

### Final Expression
Thus, the expression for the electric field is:
\[ \vec{E} = 60 \sin(0.5 \times 10^{3} x + 1.5 \times 10^{11} t) \hat{j} \text{ volts/m} \]

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