Example 39.3. Mutual Inductance of a Solenoid Surrounded by a Loop.
A solenoid loop \(C_1\) of radius \(a\) consists of \(n\) circular loops per unit length as shown in Figure 39.4.
Another circuit \(C_2\) of a circular loop of radius \(R\) surrounds the solenoid such that the solenoid is at the axis of the second loop. Assume that the length of the solenoid is much greater than the radius of the solenoid so that we can use the infinite solenoid approximation.
(a) Find the mutual inductance of the two loops from the flux of current in \(C_1\) through the area of a surface attached to the circuit \(C_2\text{.}\) Assume \(a \ll R\text{.}\)
(b) (Calculus) Find the mutual inductance of the two loops from the flux of current in \(C_2\) through the area of a surface attached to the circuit \(C_1\text{,}\) and demonstrate the two methods give rise to the same formula for the mutual inductance. Assume \(a \ll R\text{.}\)
Answer.
\(\pi \mu_0 n a^2\text{.}\)
Solution 1. (a)
(a) Calculation of mutual inductance by using current through the long solenoid.
We need magnetic flux through one circuit by current in the other circuits. Since the magnetic field of a long solenoid is easy to work with, we will think of an arbitrary current \(I\) in the solenoid as shown in Figure 39.5. This \(I\) will cancel out in the end, so we do not worry about its value.
Since, the solenoid is very long, the magnetic field is significant only inside the space of the solenoid. We use a cylindrical coordinate with the \(z\)-axis along the axis of the solenoid and the circular loop in the \(xy\)-plane. Let \(s\) be the distance of a point from the axis, i.e. the radial distance in the \(xy\)-plane.
The direction of the magnetic field for the current direction chosen in the figure is towards the positive \(z\) axis and has magnitude given by the following step function.
\begin{equation*}
B_{\text{solenoid}} =
\begin{cases}
\mu_0 n I \amp\ \ s \lt a\\
0 \amp\ \ s \ge a
\end{cases}
\end{equation*}
We attach a circular surface to the circular wire \(C_2\) to find the flux through \(C_2\text{.}\) Let \(\hat u_z\) be the normal to the surface. Since the magnetic field is zero outside the solenoid, the magnetic flux through \(C_2\) comes only from the field in the area of the circular loop that is within the cross-section of the solenoid. This gives the following for the magnetic flux \(\Phi_2\) through \(C_2\text{.}\)
\begin{equation*}
\Phi_2 = \mu_0 n I\times \pi a^2.
\end{equation*}
Dividing out the current gives the mutual inductance between the two circuits.
\begin{equation*}
\mathcal{M} = \pi \mu_0 n a^2.
\end{equation*}
Although, this formula does not look like it is symmetric in switiching properties of circuits 1 and 2. In (b), you will see that you get exactly the same expression when you switch the current on in the loop and compute the flux in the solenoid.
Solution 2. (b)
(b) Calculation of mutual inductance by using current through the circular loop.
Now, we imagine passing a current \(I\) through the loop around the solenoid so that a current \(I\) as shown in Figure 39.6. Magnetic field of this current will pass everwhere, and in particular, through a surface attached to the wires of the solenoid. We will find a formula for this flux by summing over the flux through each ring of the solenoid.
In the approximation \(a\lt \lt R\text{,}\) the magnetic field at a point in the solenoid will be approximately the same as the value at the center of solenoid coil. Since the magnetic field of the current in the outer ring depends on the distance \(|z|\) from the center of the ring, we can set up magnetic flux \(d\Phi_1\) through loops of the solenoid between \(z\) and \(z+dz\text{.}\)
Since, there are \(n dz\) loops of solenoid within a distance \(dz\text{,}\) the flux through the surface attached to this part of the solenoid will have \(n dz\) times flux through one loop at \(z\text{.}\) Therefore, \(d\Phi_1\) is given as
\begin{equation}
d\Phi_1 = B_z^{\text{Loop}}\pi a^2ndz, \tag{39.6}
\end{equation}
where \(B_z^{\text{Loop}}\) is the \(z\)-component of the magnetic field of the current in the outer loop, which has been worked out in the section on Biot-Savart law.
\begin{equation}
B_z^{\text{Loop}} = \frac{\mu_0}{2}\frac{IR^2}{\left(R^2+z^2\right)^{3/2}}\ \ \ \text{(on axis.)} \tag{39.7}
\end{equation}
Putting \(B_z^{\text{Loop}}\) from Eq. (39.7) into Eq. (39.6) and integrating from \(z=-\infty\) to \(z=\infty\) gives the following.
\begin{equation*}
\Phi_1 = \pi\mu_0 n a^2 I.
\end{equation*}
Dividing by current gives the mutual inductance
\begin{equation*}
\mathcal{M}^{\prime} = \pi\mu_0 n a^2,
\end{equation*}
which is the same as the formula of mutual inductance found by sending a current through the solenoid instead of through the outer circular wire.
