1. Condition for suspension

The wire is suspended horizontally in air, so the magnetic force balances its weight:

$F_B = mg$

For a straight current-carrying wire perpendicular to magnetic field,

$F_B = BIL$

So,

$BIL = mg$

Hence,

$B = \frac{mg}{IL}$

2. Find the length of the wire using resistance formula

Given:

• Resistance, $R = 1\,\Omega$
• Resistivity, $\rho = 2\times 10^{-6}\,\Omega\text{m}$
• Area, $A = 10\,\text{mm}^2 = 10\times 10^{-6}\,\text{m}^2 = 10^{-5}\,\text{m}^2$

Using

$R = \frac{\rho L}{A}$

So,

$L = \frac{RA}{\rho}$

Substitute values:

$L = \frac{1\times 10^{-5}}{2\times 10^{-6}} = 5\,\text{m}$

3. Compute weight of the wire

Mass:

$m = 500\,\text{g} = 0.5\,\text{kg}$

Thus,

$mg = 0.5\times 10 = 5\,\text{N}$

4. Compute magnetic field

Given current $I = 2\,\text{A}$ and length $L=5\,\text{m}$.

Using

$B = \frac{mg}{IL} = \frac{5}{2\times 5} = 0.5\,\text{T}$

Now,

$0.5\,\text{T} = 5\times 10^{-1}\,\text{T}$

So the required number is:

$\boxed{5}$