Hi there!
For a straight wire, we can use Ampère's Law to find its magnetic field at various distances from the wire.
Using the equation:
[tex]\oint B\cdot dl = B \cdot l = \mu_0 i[/tex]
B = Magnetic field strength (T)
l = length of path of integration (m)
μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)
i = Enclosed current (2.24 A)
This is a dot-product, so the cosine of the angle between the magnetic field and the path of integration is considered. However, since we are always tangential to the magnetic field, cos(0) = 1. We can simplify to B * l.
The length of the path of integration is equivalent to the circumference of a circle produced with a radius 'r' as a straight, long wire creates circular magnetic fields around the wire.
Therefore:
[tex]C = 2\pi r\\\\B \cdot 2\pi r = \mu_0 i[/tex]
Solve for 'B'.
[tex]B = \frac{\mu_0i}{2\pi r}[/tex]
1.
Plug in the given values and solve for the strength of the magnetic field at r = 0.022m.
[tex]B = \frac{(4\pi \times 10^{-7})(2.24)}{2\pi (0.022)} = \boxed{20.364 \mu T}[/tex]
Using magnetic hand rules, a left-flowing (-x axis) current will result in a magnetic field of this strength INTO THE PAGE (or +y-axis if we assign the +/- z-axis to be up/down respectively) above the wire.
2.
Since B ∝ 1/R, if we double 'R', B will be halved.
Therefore:
[tex]B_{2R} = \frac{B_R}{2} = \frac{20.364 \mu T}{2} = \boxed{10.182 \mu T}[/tex]
3.
The same logic applies. If we increase 'R' by 10x, B will decrease by 10x.
[tex]B_{10R} = \frac{B_R}{10} = \frac{20.364\mu T}{10} = \boxed{2.0364 \mu T}[/tex]
However, since this point is BELOW the wire, the direction of the magnetic field differs. Using hand rules, the field would point OUT OF THE PAGE, or to the -y-axis.
