Question

A wire of resistance 4 \(\Omega\) is used to make a coil of radius 7 cm. The wire has a diameter of 1.4 mm and the resistivity of its material is 2 x 10\(^{-7}\) \(\Omega\) m. The number of turns in the coil will be

• A. 70
• B. 40
• C. 140
• D. 20

Hint:

In such problems, break down the question into smaller parts. First, deal with the wire's intrinsic properties (resistance, resistivity, dimensions) to find the total length. Then, use that length in the context of the coil's geometry to find the required quantity (number of turns).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a multi-step problem that connects the electrical properties of a wire (resistance, resistivity) with its geometrical properties (length, area) and how it's shaped into a coil (number of turns, radius). We first need to find the total length of the wire using the resistance formula and then use that length to find the number of turns in the coil.

Step 2: Key Formula or Approach:
1. Resistance of a wire: \(R = \rho \frac{L}{A}\), where \(R\) is resistance, \(\rho\) is resistivity, \(L\) is length, and \(A\) is the cross-sectional area. 2. Cross-sectional area of the wire: \(A = \pi r_{wire}^2\), where \(r_{wire}\) is the radius of the wire. 3. Total length of the wire in the coil: \(L = n \times (\text{circumference of one turn})\), where \(n\) is the number of turns and the circumference is \(2\pi r_{coil}\).

Step 3: Detailed Explanation:
Given data:
Total resistance, \(R = 4 \, \Omega\).
Coil radius, \(r_{coil} = 7 \, \text{cm} = 0.07 \, \text{m}\).
Wire diameter, \(d_{wire} = 1.4 \, \text{mm}\), so wire radius, \(r_{wire} = 0.7 \, \text{mm} = 0.7 \times 10^{-3} \, \text{m}\).
Resistivity, \(\rho = 2 \times 10^{-7} \, \Omega \cdot \text{m}\).
Calculation:
Part 1: Find the total length (L) of the wire.
First, calculate the cross-sectional area (\(A\)) of the wire: \[ A = \pi r_{wire}^2 = \pi (0.7 \times 10^{-3} \, \text{m})^2 = \pi (0.49 \times 10^{-6} \, \text{m}^2) \] Now, use the resistance formula to find the length \(L\): \[ R = \rho \frac{L}{A} \implies L = \frac{R A}{\rho} \] \[ L = \frac{(4 \, \Omega) \times (\pi \times 0.49 \times 10^{-6} \, \text{m}^2)}{2 \times 10^{-7} \, \Omega \cdot \text{m}} \] \[ L = \frac{4 \pi \times 0.49 \times 10^{-6}}{2 \times 10^{-7}} = 2 \pi \times 0.49 \times 10 = 9.8\pi \, \text{m} \] Part 2: Find the number of turns (n).
The length of the wire required for one turn is the circumference of the coil: \[ \text{Circumference} = 2 \pi r_{coil} = 2 \pi (0.07 \, \text{m}) = 0.14\pi \, \text{m} \] The total number of turns \(n\) is the total length of the wire divided by the length of one turn: \[ n = \frac{\text{Total Length}}{\text{Length of one turn}} = \frac{L}{2 \pi r_{coil}} \] \[ n = \frac{9.8\pi \, \text{m}}{0.14\pi \, \text{m}} = \frac{9.8}{0.14} = \frac{980}{14} = 70 \]

Step 4: Final Answer:
The number of turns in the coil will be 70.