A coil of 100 turns and area 5 square centimetre is placed in a magnetic field B = 0.2 T . The normal to the plane of the coil makes an angle of `60^(@)` with the direction of the magnetic field. The magnetic flux linked with the coil is

Options

A. `5 xx 10^(-3) Wb`
B. `5 xx 10^(-5) Wb`
C. `10^(-2) Wb`
D. `10^(-4) Wb`

Answer

To find the magnetic flux linked with the coil, we can follow these steps:

### Step 1: Understand the formula for magnetic flux
The magnetic flux (Φ) linked with a coil is given by the formula:
\[ \Phi = N \cdot B \cdot A \cdot \cos(\theta) \]
where:
- \(N\) = number of turns in the coil
- \(B\) = magnetic field strength (in Tesla)
- \(A\) = area of the coil (in square meters)
- \(\theta\) = angle between the normal to the coil and the magnetic field (in degrees)

### Step 2: Convert the area from square centimeters to square meters
Given the area \(A = 5 \, \text{cm}^2\):
\[ A = 5 \, \text{cm}^2 = 5 \times 10^{-4} \, \text{m}^2 \]

### Step 3: Identify the given values
From the problem, we have:
- \(N = 100\) turns
- \(B = 0.2 \, \text{T}\)
- \(\theta = 60^\circ\)

### Step 4: Calculate \(\cos(\theta)\)
Calculate \(\cos(60^\circ)\):
\[ \cos(60^\circ) = \frac{1}{2} \]

### Step 5: Substitute the values into the formula
Now, substitute the values into the formula for magnetic flux:
\[ \Phi = 100 \cdot 0.2 \cdot (5 \times 10^{-4}) \cdot \cos(60^\circ) \]
\[ \Phi = 100 \cdot 0.2 \cdot (5 \times 10^{-4}) \cdot \frac{1}{2} \]

### Step 6: Simplify the expression
Now, simplify the expression step by step:
\[ \Phi = 100 \cdot 0.2 \cdot 5 \times 10^{-4} \cdot \frac{1}{2} \]
\[ = 100 \cdot 0.1 \cdot 5 \times 10^{-4} \]
\[ = 10 \cdot 5 \times 10^{-4} \]
\[ = 50 \times 10^{-4} \]
\[ = 5 \times 10^{-3} \, \text{Wb} \]

### Step 7: Final Result
Thus, the magnetic flux linked with the coil is:
\[ \Phi = 5 \times 10^{-3} \, \text{Wb} \]