Five very long, straight insulated wires are closely bound together to form a small cable. Currents carried by the wires are: `I_1=20A, I_2=-6A, I_3=12A, I_4=-7A, I_5=18A`. (Negative currents are opposite in direction to the positve.) The magnetic field induction at a distance of 10cm from the cable is (current enters at A and leaves at B and C as shown)

Options

A. `5muT`
B. `15muT`
C. `74muT`
D. `128muT`

Explanation

To solve the problem of finding the magnetic field induction at a distance of 10 cm from a cable formed by five long, straight insulated wires carrying different currents, we will use Ampere's circuital law. Here is the step-by-step solution:

### Step 1: Identify the currents
The currents carried by the wires are:
- \( I_1 = 20 \, \text{A} \) (positive direction)
- \( I_2 = -6 \, \text{A} \) (opposite direction)
- \( I_3 = 12 \, \text{A} \) (positive direction)
- \( I_4 = -7 \, \text{A} \) (opposite direction)
- \( I_5 = 18 \, \text{A} \) (positive direction)

### Step 2: Calculate the net current
To find the net current \( I_{\text{net}} \), we sum the individual currents:
\[ I_{\text{net}} = I_1 + I_2 + I_3 + I_4 + I_5 \]
Substituting the values:
\[ I_{\text{net}} = 20 - 6 + 12 - 7 + 18 = 37 \, \text{A} \]

### Step 3: Apply Ampere's Circuital Law
According to Ampere's circuital law:
\[ \oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I_{\text{net}} \]
For a circular path of radius \( r \), the left side becomes:
\[ B \cdot 2\pi r \]
Thus, we have:
\[ B \cdot 2\pi r = \mu_0 I_{\text{net}} \]

### Step 4: Substitute the values
We know:
- \( r = 10 \, \text{cm} = 0.1 \, \text{m} \)
- \( \mu_0 = 4\pi \times 10^{-7} \, \text{T m/A} \)

Substituting these values into the equation:
\[ B \cdot 2\pi (0.1) = 4\pi \times 10^{-7} \times 37 \]

### Step 5: Simplify the equation
Cancelling \( 2\pi \) from both sides:
\[ B \cdot 0.2 = 4 \times 10^{-7} \times 37 \]
\[ B = \frac{4 \times 10^{-7} \times 37}{0.2} \]

### Step 6: Calculate \( B \)
Calculating the right-hand side:
\[ B = \frac{148 \times 10^{-7}}{0.2} = 740 \times 10^{-7} \, \text{T} \]
\[ B = 74 \times 10^{-6} \, \text{T} = 74 \, \mu\text{T} \]

### Final Answer
The magnetic field induction at a distance of 10 cm from the cable is:
\[ B = 74 \, \mu\text{T} \]