The magnetic induction at apoint P which is at the distance 4 cm from a long current carrying wire is `10^(-3) T`. The field of induction at a distance 12 cm from the current will be

Options

A. `3.33xx10^(-4)T`
B. `1.11xx10^(-4)T`
C. `3xx10^(-3)T`
D. `9xx10^(-3)T`

Answer

To solve the problem, we will use the relationship between the magnetic induction (B) and the distance (r) from a long straight current-carrying wire. The magnetic field produced by a long straight wire is inversely proportional to the distance from the wire.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Magnetic induction at distance \( R_1 = 4 \, \text{cm} \) is \( B_1 = 10^{-3} \, \text{T} \).
- We need to find the magnetic induction \( B_2 \) at distance \( R_2 = 12 \, \text{cm} \).

2. **Use the Relationship Between Magnetic Field and Distance:**
- The magnetic field \( B \) is given by the formula:
\[ B \propto \frac{1}{r} \]
- This implies:
\[ B_1 R_1 = B_2 R_2 \]
- Where \( R_1 \) and \( R_2 \) are the distances from the wire.

3. **Substitute the Known Values:**
- Substitute \( B_1 = 10^{-3} \, \text{T} \), \( R_1 = 4 \, \text{cm} \), and \( R_2 = 12 \, \text{cm} \) into the equation:
\[ (10^{-3}) \times 4 = B_2 \times 12 \]

4. **Solve for \( B_2 \):**
- Rearranging the equation gives:
\[ B_2 = \frac{(10^{-3}) \times 4}{12} \]
- Simplifying this:
\[ B_2 = \frac{4 \times 10^{-3}}{12} = \frac{1}{3} \times 10^{-3} \, \text{T} \]

5. **Convert to a More Convenient Form:**
- To express \( B_2 \) in terms of \( 10^{-4} \):
\[ B_2 = \frac{1}{3} \times 10^{-3} = \frac{10}{3} \times 10^{-4} \, \text{T} \]
- Calculating \( \frac{10}{3} \):
\[ \frac{10}{3} \approx 3.33 \]
- Therefore:
\[ B_2 \approx 3.33 \times 10^{-4} \, \text{T} \]

### Final Answer:
The magnetic induction at a distance of 12 cm from the current-carrying wire is approximately \( 3.33 \times 10^{-4} \, \text{T} \).

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