Question 3
Solution
This problem can be solved using the principle of superposition for current densities. The cavity can be modeled as the superposition of a full cylinder carrying current density \(+\vec{J}\) and a second cylinder (occupying the position of the second circle) carrying current density \(-\vec{J}\).
Magnetic Field inside a cylinder:
Inside a cylinder with current density \(\vec{J}\), the magnetic field at position \(\vec{r}\) from the axis is given by:
$$ \vec{B} = \frac{\mu_0}{2} (\vec{J} \times \vec{r}) $$Superposition:
Let the centers of the two circles be \(O_1\) and \(O_2\). The field at a point \(P\) is:
$$ \vec{B}_{net} = \vec{B}_1 + \vec{B}_2 $$ $$ \vec{B}_{net} = \frac{\mu_0}{2} (\vec{J} \times \vec{r}_1) + \frac{\mu_0}{2} (-\vec{J} \times \vec{r}_2) $$ $$ \vec{B}_{net} = \frac{\mu_0}{2} \vec{J} \times (\vec{r}_1 – \vec{r}_2) $$From vector geometry, \(\vec{r}_1 – \vec{r}_2\) is the vector connecting the two centers, \(\vec{a}\) (or \(\vec{d}\)). This vector is constant and points along the x-axis.
Result:
Since \(\vec{J}\) is in the \(z\)-direction and \(\vec{a}\) is in the \(x\)-direction:
$$ \vec{B}_{net} = \frac{\mu_0}{2} (J\hat{k} \times a\hat{i}) = \frac{\mu_0 J a}{2} \hat{j} $$The field is uniform, points in the positive \(y\)-direction, and has a magnitude of \(\frac{1}{2}\mu_0 J a\).
Correct Answer: (c)