Question
Using Ampere’s law, derive an expression for the magnetic induction inside an ideal toroid carrying a steady current.
✓
Answer
An ideal toroid consists of a long conducting wire wound tightly around a torus made of a non $-$ conducting material.
When a steady current is passed through it, the magnetic induction $\vec{B}$ in the interior of the toroid is tangent to any circle concentric with $y$ the axis of the toroid and has the same value on this circle.
Suppose the toroid has $N$ turns of wire and $I$ is the current in its coil.
As our Amperian loop, we choose a circle of radius $r$ concentric with the axis of the toroid, as shown in figure.
Since $\vec{B}$ has the same value on this circle and is tangential to it, we go around this path in the direction of $\vec{B}$ so that $\vec{B}$ and $\overrightarrow{d l}$ are parallel.
Then, the line integral of the magnetic induction around the Amperian loop is
The net current enclosed by the Amperean loop is
$= I \times N = N \mid$
By Ampere's law,
$\oint \vec{B} \cdot \overrightarrow{d l}=\left.\mu_0\right|_{\text {encl }} \ ($in free space$)$
Therefore, from Eqs. $(1)$ and $(2),$
$B (2 \pi r)=\mu_0 N$
$\therefore B =\frac{\mu_0}{2 \pi} \frac{N I}{r} .$
This is the required expression.
When a steady current is passed through it, the magnetic induction $\vec{B}$ in the interior of the toroid is tangent to any circle concentric with $y$ the axis of the toroid and has the same value on this circle.
Suppose the toroid has $N$ turns of wire and $I$ is the current in its coil.
As our Amperian loop, we choose a circle of radius $r$ concentric with the axis of the toroid, as shown in figure.
Since $\vec{B}$ has the same value on this circle and is tangential to it, we go around this path in the direction of $\vec{B}$ so that $\vec{B}$ and $\overrightarrow{d l}$ are parallel.
Then, the line integral of the magnetic induction around the Amperian loop is
The net current enclosed by the Amperean loop is
$= I \times N = N \mid$
By Ampere's law,
$\oint \vec{B} \cdot \overrightarrow{d l}=\left.\mu_0\right|_{\text {encl }} \ ($in free space$)$
Therefore, from Eqs. $(1)$ and $(2),$
$B (2 \pi r)=\mu_0 N$
$\therefore B =\frac{\mu_0}{2 \pi} \frac{N I}{r} .$
This is the required expression.
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