A current carrying circular loop of radius \(2\) cm with unit normal \(\hat{n} = \dfrac{\hat{k}+\hat{i}}{\sqrt{2}}\) is placed in a magnetic field, \(\vec{B} = B_0(3\hat{i}+2\hat{k})\). If \(B_0 = 4\times 10^{-3}\) T and current \(I=100\sqrt{2}\) A, the torque experienced by the loop is ________ Wb·A. (\(\pi=3.14\))

Explanation:
1. Magnetic Dipole Moment (\( \vec{M} \)): The magnetic dipole moment of a current-carrying loop is given by the formula:
\( \vec{M} = I \cdot A \cdot \hat{n} \)

2. Area of the Circular Loop (\( A \)):
Given radius, \( r = 2\text{ cm} = 2 \times 10^{-2}\text{ m} \)
\( A = \pi r^2 = \pi \left(2 \times 10^{-2}\text{ m}\right)^2 = 4\pi \times 10^{-4}\text{ m}^2 \)

3. Calculating \( \vec{M} \):
Substitute \( I = 100\sqrt{2}\text{ A} \) and \( \hat{n} = \frac{\hat{i} + \hat{k}}{\sqrt{2}} \):
\( \vec{M} = \left(100\sqrt{2}\right) \times \left(4\pi \times 10^{-4}\right) \times \left(\frac{\hat{i} + \hat{k}}{\sqrt{2}}\right) \)
\( \vec{M} = 4\pi \times 10^{-2}(\hat{i} + \hat{k})\text{ A}\cdot\text{m}^2 \)

4. Torque (\( \vec{\tau} \)): The torque experienced by a magnetic dipole in a uniform magnetic field is given by:
\( \vec{\tau} = \vec{M} \times \vec{B} \)
Given, \( \vec{B} = 4 \times 10^{-3}(3\hat{i} + 2\hat{k})\text{ T} \)

Substituting the values:
\( \vec{\tau} = \left[4\pi \times 10^{-2}(\hat{i} + \hat{k})\right] \times \left[4 \times 10^{-3}(3\hat{i} + 2\hat{k})\right] \)
\( \vec{\tau} = 16\pi \times 10^{-5} \left[ (\hat{i} + \hat{k}) \times (3\hat{i} + 2\hat{k}) \right] \)

5. Evaluating Cross Product:
Using the cross product rules (\( \hat{i}\times\hat{i}=0 \), \( \hat{i}\times\hat{k}=-\hat{j} \), \( \hat{k}\times\hat{i}=\hat{j} \), \( \hat{k}\times\hat{k}=0 \)):
\( (\hat{i} + \hat{k}) \times (3\hat{i} + 2\hat{k}) = 2(\hat{i} \times \hat{k}) + 3(\hat{k} \times \hat{i}) = -2\hat{j} + 3\hat{j} = \hat{j} \)

Therefore:
\( \vec{\tau} = 16\pi \times 10^{-5}\hat{j} \)

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

Hence, Option D is the correct choice.