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The problem involves comparing the magnetic field produced by a coil in two cases. Let’s analyze the given situation step by step:

### Given:
- A coil of one turn is made from a certain length of wire.
- The same length of wire is used to make two turns of a coil in the second case.
- The same current is passed through the coil in both cases.
- We need to determine the ratio of the magnetic field in the first case (one turn) to the second case (two turns).

### Formula for the Magnetic Field:
The magnetic field \( B \) at the center of a circular coil is given by:
\[
B = \frac{\mu_0 I N}{2R}
\]
Where:
- \( \mu_0 \) is the permeability of free space (a constant).
- \( I \) is the current flowing through the coil.
- \( N \) is the number of turns of the coil.
- \( R \) is the radius of the coil.

### Step 1: Coil of One Turn
In the first case, the coil has one turn (\( N_1 = 1 \)) and a certain radius (\( R_1 \)). The magnetic field is:
\[
B_1 = \frac{\mu_0 I N_1}{2R_1} = \frac{\mu_0 I}{2R_1}
\]

### Step 2: Coil of Two Turns
In the second case, the same length of wire is used to create two turns (\( N_2 = 2 \)). Since the wire length is the same, the radius of each turn must be smaller. The total length of wire for the coil is proportional to the circumference, so for two turns, the radius must be halved:
\[
R_2 = \frac{R_1}{2}
\]
Now, the magnetic field for the two-turn coil is:
\[
B_2 = \frac{\mu_0 I N_2}{2R_2} = \frac{\mu_0 I \cdot 2}{2 \cdot \frac{R_1}{2}} = \frac{\mu_0 I}{R_1}
\]

### Step 3: Ratio of Magnetic Fields
Now, we compute the ratio of the magnetic field in the first case to the second case:
\[
\frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{2R_1}}{\frac{\mu_0 I}{R_1}} = \frac{1}{2}
\]

Thus, the ratio of the magnetic field in the first case to the second case is **1:2**.

### Final Answer:
The correct option is **(D) 1:2**.

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Would you like further details on any step, or do you have any specific questions?

Here are 5 follow-up questions for further exploration:
1. What happens to the magnetic field if the number of turns is increased to 3 using the same wire length?
2. How does the radius of the coil affect the strength of the magnetic field?
3. Why does the magnetic field increase when the number of turns increases, even though the radius decreases?
4. How would the result change if the current flowing through the coil were doubled?
5. What is the significance of the permeability constant \( \mu_0 \) in calculating the magnetic field?

**Tip:** The magnetic field strength is directly proportional to the number of turns, but inversely proportional to the radius of the coil.

A coil of one turn is made of wire of a certain length, and then from the same length of the coil, two turns are made. If the same current is passed in both cases, then the ratio between the magnetic field in the first case to the second case will be:

Explore how to calculate the ratio of the magnetic field between a one-turn coil and a two-turn coil using the same wire length. This problem uses formulas related to electromagnetism, such as the magnetic field of a coil, and is ideal for students in Grades 11-12.

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