A current carrying circular loop of radius 2 cm with unit normal n...

The magnetic dipole moment of the current carrying loop is given by $\vec{M} = I A \hat{n}$.

Given radius $r = 2$ cm $= 2 \times 10^{-2}$ m, the area of the loop is:
$A = \pi r^2 = \pi (2 \times 10^{-2})^2 = 4\pi \times 10^{-4}$ m$^2$

Substituting the given values $I = 100\sqrt{2}$ A and $\hat{n} = \dfrac{\hat{i}+\hat{k}}{\sqrt{2}}$:
$\vec{M} = (100\sqrt{2}) \times (4\pi \times 10^{-4}) \times \left(\dfrac{\hat{i}+\hat{k}}{\sqrt{2}}\right)$
$\vec{M} = 4\pi \times 10^{-2} (\hat{i}+\hat{k})$ A$\cdot$m$^2$

The torque experienced by the loop in the magnetic field is $\vec{\tau} = \vec{M} \times \vec{B}$.
Given $\vec{B} = 4 \times 10^{-3} (3\hat{i}+2\hat{k})$ T, we have:
$\vec{\tau} = [4\pi \times 10^{-2} (\hat{i}+\hat{k})] \times [4 \times 10^{-3} (3\hat{i}+2\hat{k})]$
$\vec{\tau} = 16\pi \times 10^{-5} [(\hat{i}+\hat{k}) \times (3\hat{i}+2\hat{k})]$

Evaluating the cross product:
$(\hat{i}+\hat{k}) \times (3\hat{i}+2\hat{k}) = (\hat{i} \times 3\hat{i}) + (\hat{i} \times 2\hat{k}) + (\hat{k} \times 3\hat{i}) + (\hat{k} \times 2\hat{k})$
$= 0 + 2(-\hat{j}) + 3(\hat{j}) + 0 = \hat{j}$

Therefore, the torque is:
$\vec{\tau} = 16\pi \times 10^{-5} \hat{j}$

Substituting $\pi = 3.14$:
$\vec{\tau} = 16 \times 3.14 \times 10^{-5} \hat{j} = 50.24 \times 10^{-5} \hat{j} = 5024 \times 10^{-7} \hat{j}$ Wb$\cdot$A