A coil one turn is made of a wire of certain lenghth and then from the same length a coil of two turns is made. If the same current is passed both the cases, then the ratio of magnetic induction at there centres will be

Options

A. `2 : 1`
B. `1 : 4`
C. `4 : 1`
D. `1 : 2`

Answer

To solve the problem, we need to find the ratio of the magnetic induction (magnetic field) at the centers of two coils made from the same length of wire but with different numbers of turns.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have two coils: one with 1 turn and another with 2 turns, made from the same length of wire.
- We need to find the ratio of the magnetic field at the center of both coils when the same current is passed through them.

2. **Magnetic Field Formula**:
- The magnetic field \( B \) at the center of a coil is given by the formula:
\[ B = \frac{\mu_0 N I}{2 \pi r} \]
where:
- \( \mu_0 \) = permeability of free space,
- \( N \) = number of turns,
- \( I \) = current flowing through the coil,
- \( r \) = radius of the coil.

3. **Case 1: One Turn Coil**:
- For the coil with 1 turn (\( N_1 = 1 \)):
\[ B_1 = \frac{\mu_0 \cdot 1 \cdot I}{2 \pi r_1} = \frac{\mu_0 I}{2 \pi r_1} \]

4. **Finding the Radius for the One Turn Coil**:
- The length of the wire used for the coil is equal to the circumference of the coil:
\[ L = 2 \pi r_1 \]
- Therefore, the radius \( r_1 \) can be expressed as:
\[ r_1 = \frac{L}{2 \pi} \]

5. **Case 2: Two Turns Coil**:
- For the coil with 2 turns (\( N_2 = 2 \)):
\[ B_2 = \frac{\mu_0 \cdot 2 \cdot I}{2 \pi r_2} = \frac{\mu_0 I}{\pi r_2} \]

6. **Finding the Radius for the Two Turns Coil**:
- The length of the wire used for the two-turn coil is still \( L \), but now it is divided into two turns:
\[ L = 2 \cdot 2 \pi r_2 \implies L = 4 \pi r_2 \]
- Therefore, the radius \( r_2 \) can be expressed as:
\[ r_2 = \frac{L}{4 \pi} \]

7. **Finding the Ratio of the Magnetic Fields**:
- Now we can substitute \( r_1 \) and \( r_2 \) back into the equations for \( B_1 \) and \( B_2 \):
\[ B_1 = \frac{\mu_0 I}{2 \pi \left(\frac{L}{2 \pi}\right)} = \frac{\mu_0 I}{L} \]
\[ B_2 = \frac{\mu_0 I}{\pi \left(\frac{L}{4 \pi}\right)} = \frac{4 \mu_0 I}{L} \]
- Now, we can find the ratio \( \frac{B_1}{B_2} \):
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 I}{L}}{\frac{4 \mu_0 I}{L}} = \frac{1}{4} \]

8. **Final Result**:
- Therefore, the ratio of the magnetic induction at the centers of the two coils is:
\[ B_1 : B_2 = 1 : 4 \]