Question

A magnetic field is given by \(\mathbf{B} = \nabla \times \mathbf{A}\) where \(\mathbf{A}\) is the magnetic vector potential. If \(\mathbf{A} = (ax^2 + by^2)\hat{i\), the corresponding current density \(\mathbf{J}\) is
(a and b are non-zero constants)}

• A. \(-\frac{1}{\mu_0}(2a + 2b)\hat{i}\)
• B. \(\frac{1}{\mu_0}(2a + 2b)\hat{i}\)
• C. \(-\frac{1}{\mu_0}(2a)\hat{i}\)
• D. \(-\frac{1}{\mu_0}(2b)\hat{i}\)

Hint:

When calculating curls, be systematic. Write down the determinant form and carefully evaluate each partial derivative. Remember that the curl is a vector operator, so the result will be a vector. The process is a chain: \(\mathbf{A} \xrightarrow{\nabla \times} \mathbf{B} \xrightarrow{\nabla \times} \mu_0 \mathbf{J}\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The relationship between magnetic vector potential \(\mathbf{A}\), magnetic field \(\mathbf{B}\), and current density \(\mathbf{J}\) is governed by Maxwell's equations. Specifically, the magnetic field is the curl of the vector potential, and for steady currents, the curl of the magnetic field is proportional to the current density (Ampere's Law).
Step 2: Key Formula or Approach:
1. First, calculate the magnetic field \(\mathbf{B}\) from the magnetic vector potential \(\mathbf{A}\) using the relation:
\[ \mathbf{B} = \nabla \times \mathbf{A} \] 2. Second, calculate the current density \(\mathbf{J}\) from the magnetic field \(\mathbf{B}\) using Ampere's Law in differential form:
\[ \nabla \times \mathbf{B} = \mu_0 \mathbf{J} \implies \mathbf{J} = \frac{1}{\mu_0} (\nabla \times \mathbf{B}) \] Step 3: Detailed Explanation:
Part 1: Calculate \(\mathbf{B}\)
Given \(\mathbf{A} = (ax^2 + by^2)\hat{i}\). In component form, \(A_x = ax^2 + by^2\), \(A_y = 0\), \(A_z = 0\).
The curl in Cartesian coordinates is given by:
\[ \nabla \times \mathbf{A} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\A_x & A_y & A_z \end{vmatrix} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ ax^2+by^2 & 0 & 0 \end{vmatrix} \] \[ \mathbf{B} = \hat{i}\left(\frac{\partial(0)}{\partial y} - \frac{\partial(0)}{\partial z}\right) - \hat{j}\left(\frac{\partial(0)}{\partial x} - \frac{\partial(ax^2+by^2)}{\partial z}\right) + \hat{k}\left(\frac{\partial(0)}{\partial x} - \frac{\partial(ax^2+by^2)}{\partial y}\right) \] \[ \mathbf{B} = \hat{i}(0-0) - \hat{j}(0-0) + \hat{k}(0 - 2by) \] \[ \mathbf{B} = -2by \hat{k} \] Part 2: Calculate \(\mathbf{J}\)
Now we find the curl of \(\mathbf{B}\). In component form, \(B_x = 0\), \(B_y = 0\), \(B_z = -2by\).
\[ \nabla \times \mathbf{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ 0 & 0 & -2by \end{vmatrix} \] \[ \nabla \times \mathbf{B} = \hat{i}\left(\frac{\partial(-2by)}{\partial y} - \frac{\partial(0)}{\partial z}\right) - \hat{j}\left(\frac{\partial(-2by)}{\partial x} - \frac{\partial(0)}{\partial z}\right) + \hat{k}\left(\frac{\partial(0)}{\partial x} - \frac{\partial(0)}{\partial y}\right) \] \[ \nabla \times \mathbf{B} = \hat{i}(-2b - 0) - \hat{j}(0-0) + \hat{k}(0-0) \] \[ \nabla \times \mathbf{B} = -2b \hat{i} \] Finally, using Ampere's law:
\[ \mathbf{J} = \frac{1}{\mu_0}(\nabla \times \mathbf{B}) = \frac{1}{\mu_0}(-2b \hat{i}) = -\frac{2b}{\mu_0} \hat{i} \] Step 4: Final Answer:
The corresponding current density is \(-\frac{1}{\mu_0}(2b)\hat{i}\). This matches option (D).