Two circular coils are made of two identicle wires of length 20 cm. One coil has number of turns 9 and the other has 3. If the same current flows through the coils then the ratio of magnetic fields of induclion at their centres is

Options

A. `1:9`
B. `9:1`
C. `3:1`
D. `1:3`

Answer

To solve the problem, we need to find the ratio of the magnetic fields of induction at the centers of two circular coils made from identical wires. Let's break down the solution step by step.

### Step 1: Understand the given data
We have two coils:
- Coil 1 has \( n_1 = 9 \) turns.
- Coil 2 has \( n_2 = 3 \) turns.
- The length of the wire used for both coils is \( L = 20 \) cm.
- The same current \( I \) flows through both coils.

### Step 2: Formula for magnetic field at the center of a circular coil
The magnetic field \( B \) at the center of a circular coil is given by the formula:
\[ B = \frac{\mu_0 n I}{2R} \]
where:
- \( \mu_0 \) is the permeability of free space,
- \( n \) is the number of turns,
- \( I \) is the current,
- \( R \) is the radius of the coil.

### Step 3: Determine the radius of each coil
Since both coils are made from identical wires and have the same length, we can find the radius of each coil. The circumference of a circular coil is given by:
\[ C = 2\pi R \]
For Coil 1:
\[ C_1 = 2\pi R_1 = \frac{L}{n_1} = \frac{20 \text{ cm}}{9} \]
Thus,
\[ R_1 = \frac{10}{9\pi} \text{ cm} \]

For Coil 2:
\[ C_2 = 2\pi R_2 = \frac{L}{n_2} = \frac{20 \text{ cm}}{3} \]
Thus,
\[ R_2 = \frac{10}{3\pi} \text{ cm} \]

### Step 4: Express the magnetic fields for both coils
Now we can express the magnetic fields for both coils using the formula:
\[ B_1 = \frac{\mu_0 n_1 I}{2R_1} \quad \text{and} \quad B_2 = \frac{\mu_0 n_2 I}{2R_2} \]

### Step 5: Find the ratio of the magnetic fields
To find the ratio \( \frac{B_1}{B_2} \):
\[ \frac{B_1}{B_2} = \frac{\frac{\mu_0 n_1 I}{2R_1}}{\frac{\mu_0 n_2 I}{2R_2}} = \frac{n_1 R_2}{n_2 R_1} \]
Substituting the values of \( n_1 \), \( n_2 \), \( R_1 \), and \( R_2 \):
\[ \frac{B_1}{B_2} = \frac{9 \cdot \frac{10}{3\pi}}{3 \cdot \frac{10}{9\pi}} = \frac{9 \cdot 10}{3 \cdot 10} \cdot \frac{9}{3} = \frac{9^2}{3^2} = \frac{81}{9} = 9 \]

### Conclusion
The ratio of the magnetic fields of induction at the centers of the two coils is:
\[ B_1 : B_2 = 9 : 1 \]

### Final Answer
The ratio of magnetic fields of induction at their centers is \( 9 : 1 \).

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