The magnetic field ($ \vec{B} $) can be derived from the vector potential ($ \vec{A} $) using the curl operation: $ \vec{B} = \nabla \times \vec{A} $ A uniform magnetic field of magnitude $ B_0 $ along the z-direction is represented as $ \vec{B} = (0, 0, B_0) $.
The curl of a vector field $ \vec{A} = (A_x, A_y, A_z) $ is given by: $ \nabla \times \vec{A} = \left( \frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z}, \frac{\partial A_x}{\partial z} - \frac{\partial A_z}{\partial x}, \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} \right) $ For the field to be $ \vec{B} = (0, 0, B_0) $, the following conditions must be met:
In all provided options, $ A_z = 0 $ and $ A_x, A_y $ are independent of $ z $. This automatically satisfies the first two conditions. Therefore, we only need to evaluate the third condition, $ \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} $, for each option.
Here, $ A_x = 0 $ and $ A_y = B_0x $. The z-component of the curl is: $ \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} = \frac{\partial (B_0x)}{\partial x} - \frac{\partial (0)}{\partial y} = B_0 - 0 = B_0 $ This expression yields $ \vec{B} = (0, 0, B_0) $.
Here, $ A_x = -B_0y $ and $ A_y = 0 $. The z-component of the curl is: $ \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} = \frac{\partial (0)}{\partial x} - \frac{\partial (-B_0y)}{\partial y} = 0 - (-B_0) = B_0 $ This expression yields $ \vec{B} = (0, 0, B_0) $.
Here, $ A_x = \frac{B_0x}{2} $ and $ A_y = \frac{B_0y}{2} $. The z-component of the curl is: $ \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} = \frac{\partial (\frac{B_0y}{2})}{\partial x} - \frac{\partial (\frac{B_0x}{2})}{\partial y} = 0 - 0 = 0 $ This expression yields $ \vec{B} = (0, 0, 0) $, which is not a uniform magnetic field of magnitude $ B_0 $ along the z-direction.
Here, $ A_x = -\frac{B_0y}{2} $ and $ A_y = \frac{B_0x}{2} $. The z-component of the curl is: $ \frac{\partial A_y}{\partial x} - \frac{\partial A_x}{\partial y} = \frac{\partial (\frac{B_0x}{2})}{\partial x} - \frac{\partial (-\frac{B_0y}{2})}{\partial y} = \frac{B_0}{2} - (-\frac{B_0}{2}) = B_0 $ This expression yields $ \vec{B} = (0, 0, B_0) $.
The expression that does not represent a uniform magnetic field $ \vec{B} = (0, 0, B_0) $ is Option 3, as it results in a zero magnetic field ($ \vec{B} = \vec{0} $).
The wire loop shown in the figure carries a steady current $I$. Each straight section of the loop has length $d$. A part of the loop lies in xy plane and the other part is tilted at $30^\circ$ with respect to the xz plane. The magnitude of the magnetic dipole moment of the loop (in appropriate units) is

An infinitely long cylinder of radius $R$ carries a frozen-in magnetization $\vec M = ke^{-s^2} \hat z$, where $k$ is a constant and $s$ is the distance from the axis of cylinder. The magnetic permeability of free space is $\mu_o$. There is no free current present anywhere. The magnetic flux density ($\vec B$) inside the cylinder is