Two long straight conductors are held parallel to each other `7 cm` apart. The conductors carry currents of `9A` and `16A` in opposite directions. The distance of neutral point from the conductor carrying `16A ` current is

Options

A. `9cm`
B. `16cm`
C. `25cm`
D. `63//25cm`

Explanation

To find the distance of the neutral point from the conductor carrying a current of 16 A, we can follow these steps:

### Step 1: Understand the setup
We have two long straight conductors that are parallel to each other, separated by a distance of 7 cm. One conductor carries a current of 9 A (let's call it Wire 1) and the other carries a current of 16 A (let's call it Wire 2). The currents are flowing in opposite directions.

### Step 2: Define the distances
Let the distance from Wire 2 (the 16 A wire) to the neutral point be \( D \). Consequently, the distance from Wire 1 (the 9 A wire) to the neutral point will be \( 7 - D \) since the total distance between the two wires is 7 cm.

### Step 3: Set up the magnetic field equations
At the neutral point, the magnetic fields due to both wires will be equal in magnitude but opposite in direction. The magnetic field \( B \) due to a long straight conductor carrying current \( I \) at a distance \( r \) is given by the formula:

\[ B = \frac{\mu_0 I}{2 \pi r} \]

Where \( \mu_0 \) is the permeability of free space.

For Wire 1 (9 A):
\[ B_1 = \frac{\mu_0 \cdot 9}{2 \pi (7 - D)} \]

For Wire 2 (16 A):
\[ B_2 = \frac{\mu_0 \cdot 16}{2 \pi D} \]

### Step 4: Set the magnetic fields equal
At the neutral point, we set \( B_1 = B_2 \):
\[ \frac{\mu_0 \cdot 9}{2 \pi (7 - D)} = \frac{\mu_0 \cdot 16}{2 \pi D} \]

### Step 5: Simplify the equation
We can cancel \( \mu_0 \) and \( 2 \pi \) from both sides:
\[ \frac{9}{7 - D} = \frac{16}{D} \]

### Step 6: Cross-multiply to solve for \( D \)
Cross-multiplying gives:
\[ 9D = 16(7 - D) \]
Expanding the right side:
\[ 9D = 112 - 16D \]

### Step 7: Combine like terms
Bringing all terms involving \( D \) to one side:
\[ 9D + 16D = 112 \]
\[ 25D = 112 \]

### Step 8: Solve for \( D \)
Dividing both sides by 25:
\[ D = \frac{112}{25} = 4.48 \text{ cm} \]

### Step 9: Calculate the distance from Wire 2
The distance from Wire 2 (16 A wire) to the neutral point is approximately \( 4.48 \) cm.

### Step 10: Calculate the distance from Wire 1
The distance from Wire 1 (9 A wire) to the neutral point is:
\[ 7 - D = 7 - 4.48 = 2.52 \text{ cm} \]

### Conclusion
The distance of the neutral point from the conductor carrying 16 A current is approximately \( 4.48 \) cm.

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