The horizontal component of earth's magnetic field is `3xx10^(-5) Wb//m^2`. The magnetic flux linked with a coil of area `1m^2` and having 5 turns, whose plane is normal to the magnetic field, will be

Options

A. `3xx10^(-5) Wb`
B. `5xx10^(-5) Wb`
C. `15xx10^(-5) Wb`
D. zero

Explanation

To solve the problem, we need to calculate the magnetic flux linked with a coil using the formula for magnetic flux. Here are the steps to find the solution:

### Step 1: Identify the given values
- The horizontal component of Earth's magnetic field (B) = \(3 \times 10^{-5} \, \text{Wb/m}^2\)
- The area of the coil (A) = \(1 \, \text{m}^2\)
- The number of turns in the coil (N) = \(5\)

### Step 2: Understand the orientation of the coil
Since the plane of the coil is normal (perpendicular) to the magnetic field, the angle (θ) between the magnetic field and the area vector of the coil is 0 degrees.

### Step 3: Use the formula for magnetic flux
The magnetic flux (Φ) linked with the coil can be calculated using the formula:
\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]
Where:
- \(N\) = Number of turns
- \(B\) = Magnetic field strength
- \(A\) = Area of the coil
- \(\theta\) = Angle between the magnetic field and the normal to the coil

### Step 4: Substitute the values into the formula
Since \(\theta = 0\), we have:
\[ \cos(0) = 1 \]
Now substituting the values:
\[ \Phi = 5 \cdot (3 \times 10^{-5}) \cdot 1 \cdot \cos(0) \]
\[ \Phi = 5 \cdot (3 \times 10^{-5}) \cdot 1 \cdot 1 \]
\[ \Phi = 15 \times 10^{-5} \, \text{Wb} \]

### Step 5: Final answer
Thus, the magnetic flux linked with the coil is:
\[ \Phi = 15 \times 10^{-5} \, \text{Wb} \]