Torque Experienced by Current Carrying Circular Loop in Magnetic Field

To find the torque experienced by the current-carrying circular loop, we use the relationship between magnetic torque $\vec{\tau}$, magnetic dipole moment $\vec{M}$, and magnetic field $\vec{B}$:

$\vec{\tau} = \vec{M} \times \vec{B}$

1. Calculate the Area of the Loop ($A$): Given the radius $r = 2\text{ cm} = 2 \times 10^{-2}\text{ m}$, and $\pi = 3.14$: $A = \pi r^2 = 3.14 \times (2 \times 10^{-2}\text{ m})^2 = 3.14 \times 4 \times 10^{-4}\text{ m}^2 = 12.56 \times 10^{-4}\text{ m}^2$

2. Calculate the Magnetic Dipole Moment ($\vec{M}$): The magnetic dipole moment is given by: $\vec{M} = I A \hat{n}$ Substituting $I = 100\sqrt{2}\text{ A}$ and $\hat{n} = \frac{\hat{i} + \hat{k}}{\sqrt{2}}$: $\vec{M} = 100\sqrt{2} \times 12.56 \times 10^{-4} \times \frac{\hat{i} + \hat{k}}{\sqrt{2}}$ $\vec{M} = 100 \times 12.56 \times 10^{-4} (\hat{i} + \hat{k}) = 1256 \times 10^{-4} (\hat{i} + \hat{k})\text{ A}\cdot\text{m}^2$

3. Calculate the Torque ($\vec{\tau}$): Given $\vec{B} = B_0 \left(3\hat{i} + 2\hat{k}\right) = 4 \times 10^{-3} \left(3\hat{i} + 2\hat{k}\right)\text{ T}$: $\vec{\tau} = \vec{M} \times \vec{B}$ $\vec{\tau} = \left[ 1256 \times 10^{-4} (\hat{i} + \hat{k}) \right] \times \left[ 4 \times 10^{-3} (3\hat{i} + 2\hat{k}) \right]$ $\vec{\tau} = (1256 \times 4 \times 10^{-7}) \left[ (\hat{i} + \hat{k}) \times (3\hat{i} + 2\hat{k}) \right]$

Now, evaluating the vector cross product: $(\hat{i} + \hat{k}) \times (3\hat{i} + 2\hat{k}) = 3(\hat{i} \times \hat{i}) + 2(\hat{i} \times \hat{k}) + 3(\hat{k} \times \hat{i}) + 2(\hat{k} \times \hat{k})$ Since $\hat{i} \times \hat{i} = 0$, $\hat{k} \times \hat{k} = 0$, $\hat{i} \times \hat{k} = -\hat{j}$, and $\hat{k} \times \hat{i} = \hat{j}$: $(\hat{i} + \hat{k}) \times (3\hat{i} + 2\hat{k}) = -2\hat{j} + 3\hat{j} = \hat{j}$

Substituting this back into the torque equation: $\vec{\tau} = 5024 \times 10^{-7} \hat{j}\text{ Wb}\cdot\text{A}$

Therefore, the correct option is D.