Ampère's Law Practice Test

A toroidal inductor in air has $N=300$ turns and carries a steady current $I=2.0,\text{A}$. The inner and outer radii are $a=4.0,\text{cm}$ and $b=9.0,\text{cm}$, respectively. Assume the magnetic field is negligible outside the windings and is approximately circular and tangent to a circle of radius $r$ inside the core region $a<r<b$. Consider an Amperian loop that is a circle of radius $r=5.0,\text{cm}$ centered on the toroid axis. According to Ampère's Law,

$$\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}},$$

and by symmetry $\oint \vec B\cdot d\vec \ell=B(2\pi r)$ while $I_{\text{enc}}=NI$ because the loop links all turns. Use $\mu_0=4\pi\times10^{-7},\text{T·m/A}$ and ignore fringing.

According to Ampère's Law, determine the magnetic field within the toroid at a radius of $5.0,\text{cm}$.

A long air-core solenoid has turn density $n=1200,\text{m}^{-1}$ and carries a steady current $I=0.80,\text{A}$. The solenoid length is much greater than its radius, so the magnetic field inside is approximately uniform and axial, and the magnetic field outside is approximately zero. Choose a rectangular Amperian loop with length $\ell$ inside the solenoid parallel to the axis and the return path outside. According to Ampère's Law,

$$\oint \vec B\cdot d\vec \ell=\mu_0 I_{\text{enc}},$$

with $\mu_0=4\pi\times10^{-7},\text{T·m/A}$. Using the symmetry assumptions, the integral reduces to $B\ell$ for the inside segment, and $I_{\text{enc}}=(n\ell)I$ because $n\ell$ turns are enclosed. Ignore fringing and assume the current is steady.

According to Ampère's Law, calculate the magnetic field inside the solenoid.