Two wires A and B are of lengths 40 cm and 30 cm. A is bent into a circle of radius r and B into an arc of radius r. A current `i_(1)` is passed through A and `i_(2)` through B. To have the same magnetic induction at the centre, the ratio of `i_(1)` : `i_(2)` is

Options

A. `3:4`
B. `3:5`
C. `2:3`
D. `4:3`

Answer

To solve the problem, we need to find the ratio of currents \( i_1 \) and \( i_2 \) such that the magnetic induction at the center of wire A (bent into a circle) and wire B (bent into an arc) is the same.

### Step-by-Step Solution:

1. **Determine the Circumference of Wire A**:
- The length of wire A is given as 40 cm. When bent into a circle, the circumference \( C \) is:
\[ C = 2\pi r = 40 \text{ cm} \]
- From this, we can find the radius \( r \):
\[ r = \frac{40}{2\pi} = \frac{20}{\pi} \text{ cm} \]

2. **Magnetic Induction at the Center for Wire A**:
- The formula for the magnetic induction \( B_1 \) at the center of a circular loop carrying current \( i_1 \) is given by:
\[ B_1 = \frac{\mu_0 i_1}{2r} \]

3. **Determine the Length of Wire B**:
- The length of wire B is given as 30 cm. Since it is bent into an arc of radius \( r \), we can express the angle \( \theta \) in radians subtended by the arc:
\[ \theta \cdot r = 30 \text{ cm} \]
- Substituting \( r \) from step 1:
\[ \theta \cdot \frac{20}{\pi} = 30 \implies \theta = \frac{30\pi}{20} = \frac{3\pi}{2} \text{ radians} \]

4. **Magnetic Induction at the Center for Wire B**:
- The formula for the magnetic induction \( B_2 \) at the center of an arc carrying current \( i_2 \) is:
\[ B_2 = \frac{\mu_0 i_2 \theta}{4\pi r} \]

5. **Set the Magnetic Inductions Equal**:
- To have the same magnetic induction at the center, we set \( B_1 = B_2 \):
\[ \frac{\mu_0 i_1}{2r} = \frac{\mu_0 i_2 \theta}{4\pi r} \]
- We can cancel \( \mu_0 \) and \( r \) from both sides:
\[ \frac{i_1}{2} = \frac{i_2 \theta}{4\pi} \]

6. **Substitute for \( \theta \)**:
- Substitute \( \theta = \frac{3\pi}{2} \):
\[ \frac{i_1}{2} = \frac{i_2 \cdot \frac{3\pi}{2}}{4\pi} \]
- Simplifying the right side:
\[ \frac{i_1}{2} = \frac{3i_2}{8} \]

7. **Solve for the Ratio \( \frac{i_1}{i_2} \)**:
- Cross-multiplying gives:
\[ 8i_1 = 6i_2 \implies \frac{i_1}{i_2} = \frac{6}{8} = \frac{3}{4} \]

### Final Result:
The ratio of currents \( i_1 : i_2 \) is:
\[ \frac{i_1}{i_2} = \frac{3}{4} \]