At what distance from a long straight wire carrying current of 12A will be the magnetic field be the equal to `3xx10^(-5) (Wb)//(m^(2))` ?

Options

A. `8xx10^(-2)m`
B. `12xx10^(-2)m`
C. `18xx10^(-2)m`
D. `24xx10^(-2)m`

Answer

To find the distance from a long straight wire carrying a current of 12 A at which the magnetic field is equal to \(3 \times 10^{-5} \, \text{Wb/m}^2\), we can use the formula for the magnetic field \(B\) around a long straight wire:

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

Where:
- \(B\) is the magnetic field,
- \(\mu_0\) is the permeability of free space (\(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\)),
- \(I\) is the current in amperes,
- \(R\) is the distance from the wire in meters.

### Step 1: Rearranging the formula to solve for \(R\)

We need to rearrange the formula to find \(R\):

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

### Step 2: Substitute the known values

We know:
- \(I = 12 \, \text{A}\)
- \(B = 3 \times 10^{-5} \, \text{Wb/m}^2\)
- \(\mu_0 = 4\pi \times 10^{-7} \, \text{T m/A}\)

Substituting these values into the equation:

\[ R = \frac{(4\pi \times 10^{-7}) \times 12}{2\pi \times (3 \times 10^{-5})} \]

### Step 3: Simplifying the equation

The \(\pi\) cancels out:

\[ R = \frac{4 \times 10^{-7} \times 12}{2 \times (3 \times 10^{-5})} \]

Calculating the numerator:

\[ 4 \times 12 = 48 \quad \Rightarrow \quad R = \frac{48 \times 10^{-7}}{6 \times 10^{-5}} \]

### Step 4: Further simplification

Now, simplify the fraction:

\[ R = \frac{48}{6} \times \frac{10^{-7}}{10^{-5}} = 8 \times 10^{-2} \]

### Step 5: Final result

Thus, the distance \(R\) is:

\[ R = 8 \times 10^{-2} \, \text{m} = 0.08 \, \text{m} \]

### Conclusion

The distance from the wire at which the magnetic field is \(3 \times 10^{-5} \, \text{Wb/m}^2\) is \(0.08 \, \text{m}\) or \(8 \, \text{cm}\).

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