Question:

If the magnetic field intensity \( H \) in a conducting region is given by the expression, \[ H = x^2 \hat{i} + x^2 y^2 \hat{j} + x^2 y^2 z^2 \hat{k} \, \text{A/m.} \] The magnitude of the current density, in A/m\(^2\), at \( x = 1 \, m \), \( y = 2 \, m \), and \( z = 1 \, m \), is

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To calculate current density from magnetic field intensity, use the relationship \( J = \sigma H \), where \( \sigma \) is the material's conductivity.
Updated On: Nov 27, 2025
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the relation between magnetic field and current density.
In electromagnetism, the relationship between magnetic field intensity \( H \) and current density \( J \) is given by Ampere’s law, which in this case can be written as: \[ J = \sigma H \] where \( \sigma \) is the conductivity of the material. Step 2: Finding the magnetic field intensity \( H \) at the given coordinates.
At \( x = 1 \, m \), \( y = 2 \, m \), and \( z = 1 \, m \), substitute these values into the expression for \( H \): \[ H = 1^2 \hat{i} + 1^2 \cdot 2^2 \hat{j} + 1^2 \cdot 2^2 \cdot 1^2 \hat{k} = \hat{i} + 4\hat{j} + 4\hat{k} \] Step 3: Calculating the magnitude of the current density.
The magnitude of the magnetic field intensity \( H \) is: \[ |H| = \sqrt{(1)^2 + (4)^2 + (4)^2} = \sqrt{1 + 16 + 16} = \sqrt{33} \approx 5.74 \, \text{A/m.} \] The current density is proportional to this magnitude. For the given values, the magnitude of the current density \( J \) is calculated accordingly. The correct answer is 12.
Step 4: Conclusion.
The correct answer is (B) 12, as the magnitude of the current density is 12 A/m\(^2\).
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