If the magnetic field in plane electromagnetic was is given by ` vecB = 3 xx 10 ^( -8 ) sin ( 1.6 xx 10 ^( 3 ) x + 48 xx 10 ^( 10 ) t ) hatj T`, then what will be expression for electric field ?

Options

A. ` vec E = ( 3 xx 10 ^( - 8 ) sin ( 1.6 xx 1 0 ^(3) x + 48 xx 10 ^( 10) t ) hati V//m) `
B. ` vec E = ( 9 sin (1.6 xx 10 ^( - 3) x + 48 xx 10 ^( 10 ) t ) hatk V//m) `
C. ` vecE = ( 3xx 10 ^( - 8 ) sin ( 1.6 xx 10 ^( 3 ) x + 48 xx 10 ^( 10 ) t ) hatj V//m) `
D. `vecE = ( 60 sin (1.6 xx 10 ^( 3 ) x + 48 xx 10 ^( 10 ) t ) hatk V //m) `

Answer

To find the expression for the electric field \( \vec{E} \) corresponding to the given magnetic field \( \vec{B} \) in an electromagnetic wave, we can follow these steps:

### Step 1: Identify the given magnetic field
The magnetic field is given as:
\[ \vec{B} = 3 \times 10^{-8} \sin(1.6 \times 10^{3} x + 48 \times 10^{10} t) \hat{j} \, \text{T} \]

### Step 2: Determine the direction of the electric field
In electromagnetic waves, the electric field \( \vec{E} \), magnetic field \( \vec{B} \), and the direction of wave propagation are mutually perpendicular. Given that \( \vec{B} \) is in the \( \hat{j} \) direction (y-direction), and assuming the wave propagates in the x-direction, the electric field \( \vec{E} \) will be in the z-direction (i.e., \( \hat{k} \)).

### Step 3: Use the relationship between electric and magnetic fields
The magnitudes of the electric field \( E_0 \) and magnetic field \( B_0 \) are related by the equation:
\[ E_0 = c B_0 \]
where \( c \) is the speed of light in vacuum (\( c \approx 3 \times 10^8 \, \text{m/s} \)).

### Step 4: Find \( B_0 \)
From the given magnetic field:
\[ B_0 = 3 \times 10^{-8} \, \text{T} \]

### Step 5: Calculate \( E_0 \)
Now, substituting \( B_0 \) into the equation for \( E_0 \):
\[ E_0 = c B_0 = (3 \times 10^8 \, \text{m/s})(3 \times 10^{-8} \, \text{T}) = 9 \, \text{V/m} \]

### Step 6: Write the expression for the electric field
The electric field \( \vec{E} \) can be expressed as:
\[ \vec{E} = E_0 \sin(\omega t + kx) \hat{k} \]
where \( \omega \) and \( k \) are the angular frequency and wave number, respectively.

### Step 7: Identify \( \omega \) and \( k \)
From the given magnetic field, we have:
\[ \omega = 48 \times 10^{10} \, \text{s}^{-1}, \quad k = 1.6 \times 10^{3} \, \text{m}^{-1} \]

### Step 8: Substitute \( E_0 \), \( \omega \), and \( k \) into the expression
Thus, the electric field can be written as:
\[ \vec{E} = 9 \sin(48 \times 10^{10} t + 1.6 \times 10^{3} x) \hat{k} \, \text{V/m} \]

### Final Expression
The final expression for the electric field is:
\[ \vec{E} = 9 \sin(1.6 \times 10^{3} x + 48 \times 10^{10} t) \hat{k} \, \text{V/m} \]

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