A wire of resistance `4 Omega` is used to wind a coil of radius 7 cm. The wire has a diameter of 1.4 mm and the specific resistance of its material is `2xx10^(-7) Omega m`. Find the number of turns in the coil.

Options

A. 50
B. 40
C. 60
D. 70

Explanation

To find the number of turns in the coil wound with a wire of resistance \(4 \, \Omega\), we will follow these steps:

### Step 1: Identify the given values
- Resistance of the wire, \( R = 4 \, \Omega \)
- Radius of the coil, \( r = 7 \, \text{cm} = 7 \times 10^{-2} \, \text{m} \)
- Diameter of the wire, \( d = 1.4 \, \text{mm} = 1.4 \times 10^{-3} \, \text{m} \)
- Specific resistance (resistivity) of the material, \( \rho = 2 \times 10^{-7} \, \Omega \cdot \text{m} \)

### Step 2: Calculate the radius of the wire
The radius of the wire \( R_w \) can be calculated from the diameter:
\[ R_w = \frac{d}{2} = \frac{1.4 \times 10^{-3}}{2} = 0.7 \times 10^{-3} \, \text{m} \]

### Step 3: Calculate the cross-sectional area of the wire
The cross-sectional area \( A \) of the wire can be calculated using the formula for the area of a circle:
\[ A = \pi R_w^2 = \pi (0.7 \times 10^{-3})^2 = \pi \times (0.49 \times 10^{-6}) \approx 1.539 \times 10^{-6} \, \text{m}^2 \]

### Step 4: Relate the total length of the wire to the number of turns
The total length \( L \) of the wire can be expressed in terms of the number of turns \( n \) and the circumference of the coil:
\[ L = n \cdot C = n \cdot (2 \pi r) = n \cdot (2 \pi \times 7 \times 10^{-2}) \]
Thus,
\[ L = n \cdot (2 \pi \times 0.07) \approx n \cdot 0.4398 \, \text{m} \]

### Step 5: Use the resistance formula
The resistance \( R \) of the wire is given by:
\[ R = \frac{\rho L}{A} \]
Substituting the values we have:
\[ 4 = \frac{(2 \times 10^{-7}) \cdot L}{A} \]
Substituting \( L \) from Step 4:
\[ 4 = \frac{(2 \times 10^{-7}) \cdot (n \cdot 0.4398)}{1.539 \times 10^{-6}} \]

### Step 6: Solve for \( n \)
Rearranging the equation:
\[ 4 \cdot 1.539 \times 10^{-6} = (2 \times 10^{-7}) \cdot (n \cdot 0.4398) \]
Calculating the left side:
\[ 6.156 \times 10^{-6} = (2 \times 10^{-7}) \cdot (n \cdot 0.4398) \]
Now, divide both sides by \( 2 \times 10^{-7} \):
\[ n \cdot 0.4398 = \frac{6.156 \times 10^{-6}}{2 \times 10^{-7}} = 30.78 \]
Finally, solve for \( n \):
\[ n = \frac{30.78}{0.4398} \approx 69.9 \approx 70 \]

### Conclusion
The number of turns in the coil is approximately \( 70 \).

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