A coil of `50` turns and `20 cm` diameter is made with a wire of `0.2 mm` diameter and resistivity `2xx10^(-6)Omega cm`.The coil is connected to a source of `EMF`` 20 V` and negligible internal resistance.
Find the current through the coil.

Explanation

To find the current through the coil, we will follow these steps:

### Step 1: Gather the given data
- Number of turns (N) = 50
- Diameter of the coil (d_c) = 20 cm = 0.2 m
- Diameter of the wire (d_w) = 0.2 mm = 0.2 x 10^(-3) m
- Resistivity (ρ) = 2 x 10^(-6) Ω cm = 2 x 10^(-8) Ω m
- EMF (V) = 20 V

### Step 2: Convert the diameter of the coil to radius
- Radius of the coil (r_c) = d_c / 2 = 0.2 m / 2 = 0.1 m

### Step 3: Convert the diameter of the wire to radius
- Radius of the wire (r_w) = d_w / 2 = (0.2 x 10^(-3) m) / 2 = 0.1 x 10^(-3) m = 1 x 10^(-4) m

### Step 4: Calculate the length of the wire (L)
- The length of the wire can be calculated using the formula for the circumference of a circle multiplied by the number of turns:
\[ L = N \times (2 \pi r_c) = 50 \times (2 \pi \times 0.1) = 50 \times 0.2\pi = 10\pi \text{ m} \]

### Step 5: Calculate the cross-sectional area of the wire (A)
- The cross-sectional area of the wire is given by:
\[ A = \pi r_w^2 = \pi (1 \times 10^{-4})^2 = \pi \times 10^{-8} \text{ m}^2 \]

### Step 6: Calculate the resistance (R) of the wire
- The resistance can be calculated using the formula:
\[ R = \frac{\rho L}{A} \]
Substituting the values:
\[ R = \frac{(2 \times 10^{-8}) \times (10\pi)}{\pi \times 10^{-8}} = \frac{2 \times 10^{-8} \times 10\pi}{\pi \times 10^{-8}} = 20 \text{ ohms} \]

### Step 7: Calculate the current (I) using Ohm's Law
- According to Ohm's Law:
\[ I = \frac{V}{R} \]
Substituting the values:
\[ I = \frac{20 \text{ V}}{20 \text{ ohms}} = 1 \text{ A} \]

### Final Answer
The current through the coil is **1 A**.

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