Ampère's Law — Question 5

Question 5

An off-axis cylindrical cavity contains a uniform field from superposed currents

An infinite cylinder carries uniform current density $\vec J=J\hat z$ except for an empty cylindrical cavity whose axis is displaced by $\vec d=d\hat x$. The cavity lies wholly inside the cylinder. Use the interior field of a uniform full cylinder, $\vec B=\mu_0\vec J\times\vec r/2$. Use vacuum permeability $\mu_0\simeq4\pi\times10^{-7}\,\mathrm{T\,m/A}$ where numerical values are needed.

1. Use superposition to find the field everywhere inside the cavity.

2. Evaluate its magnitude and direction for J=$2.00\times10^6$ A/m squared and d=3.00 mm. Does the cavity radius enter?

Question 5 - Solution

An off-axis cylindrical cavity contains a uniform field from superposed currents

(a) Method and calculation

$$\vec r\prime=\vec r-\vec d,\quad\vec B=\frac{\mu_0}{2}\vec J\times\vec r+\frac{\mu_0}{2}(-\vec J)\times(\vec r-\vec d).$$$$\boxed{\vec B_{\mathrm{cavity}}=\frac{\mu_0}{2}\vec J\times\vec d=\frac{\mu_0Jd}{2}\hat y}.$$

(b) Interpretation and further calculation

$$B=\frac{4\pi\times10^{-7}}2(2.00\times10^6)(.00300)=\boxed{3.76991\,\mathrm{mT}}.$$$$\partial\vec B/\partial\vec r=0\quad\text{inside the cavity};\quad\vec B\text{ is independent of cavity radius}.$$

Physical check

The cancellation removes all position dependence within the cavity. This result relies on uniform remaining current density, not uniform total current divided by the full area.