A heater coil is to be constructed with a nichrome wire `(rho = 1.0 xx 10^(-6) Omegam)` which can operate at `500 W` when connected to a `250 V` supply. (a) What would be the resistance of the coil? (b) If the cross-sectional area of the wire is `0.5 mm^2`, what length of the wire will be needed? (c) If the radius of each turn is `4.0 mm`, how many turns will be there in the coil?

Explanation

Let's solve the problem step by step:

### Given Data:
- Resistivity of nichrome wire, \( \rho = 1.0 \times 10^{-6} \, \Omega \, m \)
- Power, \( P = 500 \, W \)
- Voltage, \( V = 250 \, V \)
- Cross-sectional area of the wire, \( A = 0.5 \, mm^2 = 0.5 \times 10^{-6} \, m^2 \)
- Radius of each turn, \( r = 4.0 \, mm = 4.0 \times 10^{-3} \, m \)

### (a) Calculate the Resistance of the Coil
Using the formula for power in terms of voltage and resistance:
\[ P = \frac{V^2}{R} \]
Rearranging gives:
\[ R = \frac{V^2}{P} \]
Substituting the values:
\[ R = \frac{(250)^2}{500} = \frac{62500}{500} = 125 \, \Omega \]

### (b) Calculate the Length of the Wire
Using the formula for resistance:
\[ R = \frac{\rho L}{A} \]
Rearranging gives:
\[ L = \frac{R \cdot A}{\rho} \]
Substituting the values (note that \( A \) must be in \( m^2 \)):
\[ L = \frac{125 \cdot (0.5 \times 10^{-6})}{1.0 \times 10^{-6}} = \frac{125 \cdot 0.5}{1.0} = 62.5 \, m \]

### (c) Calculate the Number of Turns in the Coil
The total length of the wire is equal to the circumference of each turn multiplied by the number of turns:
\[ L = 2 \pi r \cdot n \]
Rearranging gives:
\[ n = \frac{L}{2 \pi r} \]
Substituting the values:
\[ n = \frac{62.5}{2 \pi (4.0 \times 10^{-3})} \]
Calculating:
\[ n = \frac{62.5}{2 \cdot 3.14 \cdot 0.004} \approx \frac{62.5}{0.02512} \approx 2487.5 \]
Rounding to the nearest whole number gives:
\[ n \approx 2500 \text{ turns} \]

### Final Answers:
(a) Resistance of the coil: \( 125 \, \Omega \)
(b) Length of the wire needed: \( 62.5 \, m \)
(c) Number of turns in the coil: \( 2500 \)