The magnetic field \( B \) around a long straight wire carrying a current \( I \) is given by Ampère's law through the formula:
\[ B = \frac{\mu_0 I}{2 \pi r} \]
where:
- \( B \) is the magnetic field,
- \( \mu_0 \) is the permeability of free space (a constant),
- \( I \) is the current through the wire,
- \( r \) is the distance from the wire.
In this case, the distance \( r = 0.01 \) m remains constant as the current changes from \( 2 \) A to \( 6 \) A.
To find out how the magnetic field changes with the increase in current, we can set up a ratio of the magnetic field at the two different currents:
\[ \frac{B_2}{B_1} = \frac{\frac{\mu_0 I_2}{2 \pi r}}{\frac{\mu_0 I_1}{2 \pi r}} = \frac{I_2}{I_1} \]
Substituting the values of the currents:
\[ \frac{B_2}{B_1} = \frac{6, \text{A}}{2, \text{A}} = 3 \]
This means the magnetic field at 6 A is three times the magnetic field at 2 A.
Therefore, the change in the magnetic field when the current increases from 2 A to 6 A is that it increases to three times its original value.
Thus, the correct answer is:
It increases to three times its original value.