## Self-inductance of Solenoid

### Task number: 1546

A long cylindrical solenoid with 100 loops per 1 cm has a radius of 1.6 cm. Assume the magnetic field inside the solenoid to be homogeneous and parallel to the axis of the solenoid.

a) What is the inductance of the solenoid per 1 meter of its length?

b) What electromotive force is induced per 1 meter of the length of the solenoid if the current change is 13 As^{-1}?

#### Hint

What is the relation for electromagnetic force induced in the coil? What quantities does the coil inductance depend on and in what manner?

#### Analysis

We will consider the number of turns of the coil to be

*N*/m. We can easily determine that if the number of turns per 1 cm is 100, then there is 10,000 turns per 1 meter. We can also imagine that it is a 1 meter long coil with 10,000 turns.a) Inductance of a coil is defined as

\[L=\frac{\Phi}{I}\]The overall magnetic flux is given by the number of turns (in our case the number of turns per 1 m) multiplied by the magnetic flux through one turn. To determine this flux we apply the relation for magnetic induction inside the solenoid.

b) A current flowing in one turn generates a magnetic flux through the turn. The flux is directly proportional to the current. All of the

*N*turns generate an overall flux*N*Φ. The electromotive force induced in the coil is determined as the overall flux per delta time. We evaluate the overall flux from the formula for coil inductance#### Solution

a) The inductance of the solenoid is defined as the constant of proportionality between the total magnetic flux through the coil turns and the current flowing in the coil:

\[N\Phi = LI\]where

\[\Phi = BS\cos\,\alpha\]*N*is the number of turns per one meter, thus*N*= 10,000. The magnetic flux Φ is the flux through one turn. Since the magnetic field is homogeneous, it is true thatThe magnetic induction

\[S=\pi R^{2}\]*B*is at all points perpendicular to the cross section of the coil, therefore cos*α*= 1, where*α*is the angle between the magnetic induction*B*and the normal to the surface, thus*α*= 0^{o}. The solenoid has a circular cross-section, therefore the area isWe substitute the formula for magnetic induction inside the coil for the magnetic induction

\[B = \mu_o NI\]*B*.After substituting all of these derived relations into the first relation, we obtain

\[N\mu_o NI\pi R^{2}=LI\]We evaluate the unknown inductance

\[L=\mu_o \pi N^{2}R^{2}\]*L*b) For the electromotive force induced in the coil the relationship

\[|U_i| =\left| \frac {\mathrm{d}(N\Phi)}{\mathrm{d} t} \right|\]applies.

For the inductance L of a one meter long coil it applies:

\[N \Phi = L I\]For the left side of the equation we substitute the formula for the coil inductance, thus the formula contains the change of current per time known from the assignment.

\[|U_i| =L \frac {\mathrm{d} I}{\mathrm{d} t}\]#### Given values and numerical insertion

\[n = 10\,000\,\mathrm{m^{-1}}\] \[r=1.6\,\mathrm{cm} = 0.016\,\mathrm{m}\] \[\frac{\mathrm{d}I}{\mathrm{d}t} = 13\,\mathrm{As^{-1}}\] \[L = \mathrm{?}\] \[U_i =\mathrm{?}\]

\[L=\mu_o \pi n^{2}R^{2}=4 \pi \cdot10^{-7}\cdot \pi \cdot 10^{8} \cdot 0.016^{2}\,\mathrm{H} = 0.1\,\mathrm{H} \] \[|U_i| =L \, \frac {d I}{d t}= 0{.}1 {\cdot} 13 \,\mathrm{V}=1{.}3\,\mathrm{V}\]#### Answer

The inductance of a one meter long solenoid is L = 0.1 H.

The solenoid induces the voltage

*U*_{i}= 1.3 V per one meter.