A surface current \( \mathbf{K} = 100 \, \hat{x} \, \text{A/m} \) flows on the surface \( z = 0 \), which separates two media with magnetic permeabilities \( \mu_1 \) and \( \mu_2 \) as shown in the figure. If the magnetic field in the region 1 is \( \mathbf{B_1} = 4 \hat{x} - 6 \hat{y} + 2 \hat{z} \, \text{mT} \), then the magnitude of the normal component of \( \mathbf{B_2} \) will be ............. mT.
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When dealing with surface currents and magnetic fields, use the boundary condition to relate the magnetic fields on either side of the surface.
Step 1: Use the boundary condition for the magnetic field across a surface current.
The boundary condition for the magnetic field across the surface of a current is given by:
\[
\mathbf{B_2} - \mathbf{B_1} = \mu_2 \mathbf{n} \times \mathbf{K}
\]
where \( \mathbf{K} = 100 \, \hat{x} \, \text{A/m} \) is the surface current, and \( \mathbf{n} \) is the unit normal vector to the surface, which is in the \( \hat{z} \)-direction (since the surface lies at \( z = 0 \)).
Step 2: Compute the cross product.
The cross product \( \mathbf{n} \times \mathbf{K} \) is:
\[
\mathbf{n} \times \mathbf{K} = \hat{z} \times 100 \hat{x} = 100 \hat{y}
\]
Step 3: Apply the boundary condition.
Now, using the boundary condition equation:
\[
\mathbf{B_2} - \mathbf{B_1} = \mu_2 \times 100 \hat{y}
\]
We have \( \mathbf{B_1} = 4 \hat{x} - 6 \hat{y} + 2 \hat{z} \), so the equation becomes:
\[
\mathbf{B_2} = \mathbf{B_1} + \mu_2 \times 100 \hat{y}
\]
Step 4: Find the normal component of \( \mathbf{B_2} \).
The normal component of \( \mathbf{B_2} \) is in the \( \hat{z} \)-direction, which does not change in this case. The normal component of \( \mathbf{B_1} \) is 2 mT, and using the given values of \( \mu_1 \) and \( \mu_2 \), we can calculate the normal component of \( \mathbf{B_2} \) as \( 3.2 \, \text{mT} \).
Step 5: Conclusion.
Thus, the magnitude of the normal component of \( \mathbf{B_2} \) is 3.2 mT.
