Question

a toroid is a solenoid in the shape of a donut (figure 1). use amperes law along the circular paths, shown dashed in the figure, to determine the magnetic field inside the toroid, where n is the total number of turns. express your answer in terms of the variables n, i, r, and the constants π and μ0. part b determine the magnetic field outside the toroid. express your answer in terms of the variables n, i, r, and the constants π and μ0.

Explanation:

Step1: Apply Ampere's law

Ampere's law is $\oint\vec{B}\cdot d\vec{l}=\mu_0I_{enclosed}$. For a toroid, consider a circular Amperian - loop of radius $r$ (where $r$ is the radius of the circular path inside the toroid) with $N$ total turns and current $I$ in each turn. The length of the Amperian - loop is $l = 2\pi r$. The total current enclosed by the loop is $I_{enclosed}=NI$. So, $\oint\vec{B}\cdot d\vec{l}=B\times2\pi r$ (since $\vec{B}$ is parallel to $d\vec{l}$ everywhere on the loop). Then $B\times2\pi r=\mu_0NI$.

Step2: Solve for $B$ inside the toroid

Rearranging the equation $B\times2\pi r=\mu_0NI$ for $B$, we get $B = \frac{\mu_0NI}{2\pi r}$. Here, if we assume the mean - radius of the toroid is $R$, then $B=\frac{\mu_0NI}{2\pi R}$.

Step3: Analyze the magnetic field outside the toroid

For a circular Amperian - loop outside the toroid, the net current enclosed by the loop is zero. According to Ampere's law $\oint\vec{B}\cdot d\vec{l}=\mu_0I_{enclosed}$, and since $I_{enclosed} = 0$, we have $B = 0$.

Answer:

Part A: $B=\frac{\mu_0NI}{2\pi R}$
Part B: $B = 0$