Question

Which one of the following is an impossible magnetic field \(\vec{B}\)?

• A. \(\vec{B} = 3z^2 \hat{x} - 2x^2 \hat{z}\)
• B. \(\vec{B} = -2xy \hat{x} + x^2 y \hat{y} + \left( \frac{2yz - x^2}{3} \right) \hat{z}\)
• C. \(\vec{B} = (xz + 4y) \hat{x} - xy^3 \hat{y} + \left( \frac{x^2z - z^2}{2} \right) \hat{z}\)
• D. \(\vec{B} = -6xz \hat{x} + 3y^2 \hat{y}\)

Hint:

For magnetic fields, always check if the divergence is zero using Gauss's law for magnetism, which is a necessary condition for the field to be valid.

Answer 1

The Correct Option is D

Step 1: Understanding the conditions for a magnetic field.
A magnetic field \(\vec{B}\) must satisfy the condition \(\nabla \cdot \vec{B} = 0\) (Gauss's law for magnetism), meaning that the divergence of the magnetic field must be zero. We will check the divergence for each option.

Step 2: Calculating the divergence for each option.
- Option (A): \(\vec{B} = 3z^2 \hat{x} - 2x^2 \hat{z}\) \(\nabla \cdot \vec{B} = \frac{\partial}{\partial x}(3z^2) + \frac{\partial}{\partial y}(0) + \frac{\partial}{\partial z}(-2x^2) = 0\). The divergence is zero, so this is a valid magnetic field.
- Option (B): \(\vec{B} = -2xy \hat{x} + x^2 y \hat{y} + \left( \frac{2yz - x^2}{3} \right) \hat{z}\) \(\nabla \cdot \vec{B} = \frac{\partial}{\partial x}(-2xy) + \frac{\partial}{\partial y}(x^2 y) + \frac{\partial}{\partial z}\left(\frac{2yz - x^2}{3}\right) = 0\). The divergence is zero, so this is a valid magnetic field.
- Option (C): \(\vec{B} = (xz + 4y) \hat{x} - xy^3 \hat{y} + \left( \frac{x^2z - z^2}{2} \right) \hat{z}\) \(\nabla \cdot \vec{B} = \frac{\partial}{\partial x}(xz + 4y) + \frac{\partial}{\partial y}(-xy^3) + \frac{\partial}{\partial z}\left(\frac{x^2z - z^2}{2}\right) = x - 3xy^2 + \frac{x^2 - 2z}{2}\). This does not give zero, so this is an impossible magnetic field.
- Option (D): \(\vec{B} = -6xz \hat{x} + 3y^2 \hat{y}\) \(\nabla \cdot \vec{B} = \frac{\partial}{\partial x}(-6xz) + \frac{\partial}{\partial y}(3y^2) = -6z + 6y\), which gives zero for specific values, so it can be a valid magnetic field.

Step 3: Conclusion.
The correct answer is (C) because the divergence does not satisfy the condition \(\nabla \cdot \vec{B} = 0\).