If the magnetic field \(\rm \overline{\mathrm{H}}=[3 x \cos \beta+6 y \sin \alpha] \overline{\mathrm{a}}_{z}\), find the current density J if fields are invariant with time.

Explanation:

Analysis of Current Density (\(\rm \bar{J}\)) from the Given Magnetic Field

Problem Statement: The given magnetic field is \(\rm \overline{\mathrm{H}}=[3 x \cos \beta+6 y \sin \alpha] \overline{\mathrm{a}}_{z}\). We are tasked with finding the current density (\(\rm \bar{J}\)) under the condition that the fields are invariant with time.

Solution Approach:

To find the current density (\(\rm \bar{J}\)), we use Maxwell's equation:

\(\rm \nabla \times \overline{\mathrm{H}} = \bar{J}\)

The magnetic field \(\rm \overline{\mathrm{H}}\) is given as:

\(\rm \overline{\mathrm{H}} = \left[3x\cos\beta + 6y\sin\alpha\right] \overline{\mathrm{a}}_{z}\)

Since \(\rm \overline{\mathrm{H}}\) is expressed in the \(\rm \overline{\mathrm{a}}_{z}\) direction, we need to calculate the curl of \(\rm \overline{\mathrm{H}}\) in Cartesian coordinates. The curl of a vector field \(\rm \overline{\mathrm{H}} = H_x \overline{\mathrm{a}}_x + H_y \overline{\mathrm{a}}_y + H_z \overline{\mathrm{a}}_z\) is given by:

\(\rm \nabla \times \overline{\mathrm{H}} = \begin{vmatrix} \overline{\mathrm{a}}_x & \overline{\mathrm{a}}_y & \overline{\mathrm{a}}_z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ H_x & H_y & H_z \end{vmatrix}\)

Here, \(\rm H_x = 0\), \(\rm H_y = 0\), and \(\rm H_z = 3x\cos\beta + 6y\sin\alpha\).

Substituting these values into the determinant:

\(\rm \nabla \times \overline{\mathrm{H}} = \begin{vmatrix} \overline{\mathrm{a}}_x & \overline{\mathrm{a}}_y & \overline{\mathrm{a}}_z \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & 0 & 3x\cos\beta + 6y\sin\alpha \end{vmatrix}\)

Expanding the determinant:

\(\rm \nabla \times \overline{\mathrm{H}} = \overline{\mathrm{a}}_x \left(\frac{\partial}{\partial y}(3x\cos\beta + 6y\sin\alpha) - \frac{\partial}{\partial z}(0)\right) - \overline{\mathrm{a}}_y \left(\frac{\partial}{\partial x}(3x\cos\beta + 6y\sin\alpha) - \frac{\partial}{\partial z}(0)\right) + \overline{\mathrm{a}}_z \left(\frac{\partial}{\partial x}(0) - \frac{\partial}{\partial y}(0)\right)\)

Simplifying term by term:

• \(\rm \overline{\mathrm{a}}_x\): The partial derivative with respect to \(y\) is \(\rm \frac{\partial}{\partial y}(3x\cos\beta + 6y\sin\alpha) = 6\sin\alpha\).
• \(\rm \overline{\mathrm{a}}_y\): The partial derivative with respect to \(x\) is \(\rm \frac{\partial}{\partial x}(3x\cos\beta + 6y\sin\alpha) = 3\cos\beta\).
• \(\rm \overline{\mathrm{a}}_z\): Both partial derivatives (\(\rm \frac{\partial}{\partial x}(0)\) and \(\rm \frac{\partial}{\partial y}(0)\)) are zero.

Thus, the curl becomes:

\(\rm \nabla \times \overline{\mathrm{H}} = 6\sin\alpha \overline{\mathrm{a}}_x - 3\cos\beta \overline{\mathrm{a}}_y\)

By Maxwell's equation (\(\rm \nabla \times \overline{\mathrm{H}} = \bar{J}\)), the current density is:

\(\rm \bar{J} = 6\sin\alpha \overline{\mathrm{a}}_x - 3\cos\beta \overline{\mathrm{a}}_y \; \mathrm{A/m^2}\)

Correct Option:

The correct option is:

Option 4: \(\rm \bar{J} = 6\sin\alpha \overline{\mathrm{a}}_x - 3\cos\beta \overline{\mathrm{a}}_y \; \mathrm{A/m^2}\)

Important Information

To further understand the analysis, let’s evaluate the other options:

Option 1: \(\rm \bar{J} = 6\sin\alpha \overline{\mathrm{a}}_x + 3\cos\beta \overline{\mathrm{a}}_y \; \mathrm{A/m^2}\)

This option incorrectly represents the direction of the current density in the \(\rm \overline{\mathrm{a}}_y\) component. The correct sign for the \(\rm \overline{\mathrm{a}}_y\) component is negative (\(-3\cos\beta\)), as determined during the curl computation.

Option 2: \(\rm \bar{J} = 6\sin\alpha \overline{\mathrm{a}}_x + 3\cos\beta \overline{\mathrm{a}}_y \; \mathrm{A/m}\)

This option has two errors: the incorrect sign for the \(\rm \overline{\mathrm{a}}_y\) component (as explained in option 1) and the incorrect units for current density. The unit of current density is \(\rm \mathrm{A/m^2}\), not \(\rm \mathrm{A/m}\).

Option 3: \(\rm \bar{J} = 6\sin\alpha \overline{\mathrm{a}}_x - 3\cos\beta \overline{\mathrm{a}}_y \; \mathrm{A/m}\)

While this option correctly represents the directions of the components, it uses the wrong unit for current density. The correct unit is \(\rm \mathrm{A/m^2}\), not \(\rm \mathrm{A/m}\).

Option 4: \(\rm \bar{J} = 6\sin\alpha \overline{\mathrm{a}}_x - 3\cos\beta \overline{\mathrm{a}}_y \; \mathrm{A/m^2}\)

This option accurately represents both the directions of the components and the correct units for current density. It matches the result obtained from the curl computation.

Conclusion:

Understanding Maxwell's curl equation and correctly applying it to the given magnetic field is essential for computing the current density. The calculation confirms that the correct current density is \(\rm \bar{J} = 6\sin\alpha \overline{\mathrm{a}}_x - 3\cos\beta \overline{\mathrm{a}}_y \; \mathrm{A/m^2}\), making option 4 the correct choice.