Question

A 36.5 mA current is carried by a uniformly wound air-core solenoid with 430 turns, a 18.5 mm diameter, and 11.5 cm length. 

(a) Compute the magnetic field inside the solenoid. 

 (b) Compute the magnetic flux through each turn. 

 (c) Compute the inductance of the solenoid. mH (d) Which of these quantities depends on the current? (Select all that apply.) 

magnetic field inside the solenoid magnetic flux through each turn inductance of the solenoid

Answer 1

Concepts and reason

The concept required to solve this problem are magnetic field inside a solenoid, magnetic flux, and self-inductance.

Initially, use the formula of magnetic field to solve for the magnetic field inside the solenoid.

Later, solve for the area and substitute in the magnetic flux equation to calculate the magnetic flux through each turn of the solenoid. Then calculate the self-inductance of the solenoid from the equation of self-inductance.

Finally, simplify the expressions of magnetic field, flux and self-inductance to identify the quantities that depends on the current in the coil.

Fundamentals

The expression for magnetic field B inside the solenoid is given as follows:

$B = {\mu _0}\frac{N}{l}i$

Here, ${\mu _0}$ is the permeability of vacuum, $N$ is the number of turns of the solenoid, $l$ is the length of the solenoid and $i$ is the current through the solenoid.

The magnetic flux through the coil is given as follows:

$\Phi = BA$

Here, B is the magnetic field, and A is the area of the coil.

The area of the circular cross section of cylinder is,

$A = \pi {r^2}$

Here, $r$ is the radius of solenoid.

The self-inductance of the solenoid is given as,

$L = \frac{{N\Phi }}{I}$

Here, ${\mu _0}$ is the permeability of vacuum, $N$ is the number of turns of the solenoid, $A$ is the area of cross section of the solenoid, and $l$ is the length of the solenoid.

(a)

Use the equation of magnetic field due to a solenoid at its center.

Substitute $4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}$ for ${\mu _0}$ , 430 for N, 11.5 cm for $l$ , and $36.5{\rm{ mA}}$ for $i$ in the equation $B = {\mu _0}\frac{N}{l}i$ and calculate the magnetic field inside the solenoid.

$\begin{array}{c}\\B = \frac{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}} \right)\left( {430} \right)}}{{\left( {11.5{\rm{ cm}}} \right)}}\left( {{\rm{36}}{\rm{.5 mA}}} \right)\\\\ = \frac{{\left( {4\pi \times {{10}^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}}} \right)\left( {430} \right)}}{{\left( {11.5{\rm{ cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}}\left( {{\rm{36}}{\rm{.5 mA}}\left( {\frac{{{{10}^{ - 3}}{\rm{ A}}}}{{1{\rm{ mA}}}}} \right)} \right)\\\\ = 1.72 \times {10^{ - 4}}{\rm{ T}}\\\end{array}$

(b)

Use the equation of magnetic flux.

$\Phi = BA$

Substitute $\pi {r^2}$ for $A$ and $\frac{d}{2}$ for $r$ in the equation $\Phi = BA$ .

$\begin{array}{c}\\\Phi = B\left( {\pi {r^2}} \right)\\\\ = B\left( {\pi {{\left( {\frac{d}{2}} \right)}^2}} \right)\\\\ = B\pi \frac{{{d^2}}}{4}\\\end{array}$

Substitute $1.72 \times {10^{ - 4}}{\rm{ T}}$ for $B$ , and $18.5{\rm{ mm}}$ for $d$ in the equation $\Phi = B\pi \frac{{{d^2}}}{4}$ .

$\begin{array}{c}\\\Phi = \left( {1.72 \times {{10}^{ - 4}}{\rm{ T}}} \right)\pi \frac{{{{\left( {18.5{\rm{ mm}}} \right)}^2}}}{4}\\\\ = \left( {1.72 \times {{10}^{ - 4}}{\rm{ T}}} \right)\pi \frac{{{{\left( {18.5{\rm{ mm}}\left( {\frac{{{{10}^{ - 3}}{\rm{ m}}}}{{1{\rm{ mm}}}}} \right)} \right)}^2}}}{4}\\\\ = 4.57 \times {10^{ - 8}}\;{\rm{T}} \cdot {{\rm{m}}^2}\\\end{array}$

(c)

Use the equation of self-inductance.

Substitute 430 for N, $4.6 \times {10^{ - 8}}\,{\rm{T}} \cdot {{\rm{m}}^2}$ for $\Phi$ , and 36.5 mA for $l$ in the equation $L = \frac{{N\Phi }}{I}$ and calculate L.

$\begin{array}{c}\\L = \frac{{\left( {430} \right)\left( {{\rm{4}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{ - 8}}{\rm{ T}} \cdot {{\rm{m}}^2}} \right)}}{{\left( {36.5{\rm{ mA}}} \right)}}\\\\ = \frac{{\left( {430} \right)\left( {{\rm{4}}{\rm{.6}} \times {\rm{1}}{{\rm{0}}^{ - 8}}{\rm{ T}} \cdot {{\rm{m}}^2}} \right)}}{{\left( {36.5{\rm{ mA}}\left( {\frac{{{{10}^{ - 3}}{\rm{ A}}}}{{1{\rm{ mA}}}}} \right)} \right)}}\\\\ = 5.42 \times {10^{ - 4}}{\rm{ H}}\left( {\frac{{{{10}^3}{\rm{ mH}}}}{{1\,{\rm{H}}}}} \right)\\\\ = 0.54{\rm{ mH}}\\\end{array}$

(d.a)

Use the equation of magnetic field inside the solenoid in the equation of magnetic flux.

Substitute ${\mu _0}\frac{N}{l}I$ for $B$ in the equation of magnetic flux $\Phi = BA$ .

$\begin{array}{c}\\\Phi = BA\\\\ = {\mu _0}\frac{N}{l}IA\\\end{array}$

Substitute ${\mu _0}\frac{N}{l}IA$ in the equation of self-inductance $L = \frac{{N\Phi }}{I}$ .

$\begin{array}{c}\\L = \frac{{N\left( {{\mu _0}\frac{N}{l}IA} \right)}}{I}\\\\ = \frac{{{N^2}{\mu _0}A}}{l}\\\end{array}$

Thus, the self-inductance does not depend on the current.

The equation of magnetic field inside the solenoid is,

$B = {\mu _0}\frac{N}{l}I$

Clearly, from the formula of the magnetic field inside the solenoid, it depends on the current.

(d.b)

Use the equation of magnetic field inside the solenoid in the equation of magnetic flux.

Substitute ${\mu _0}\frac{N}{l}I$ for $B$ in the equation of magnetic flux $\Phi = BA$ .

$\begin{array}{c}\\\Phi = BA\\\\ = {\mu _0}\frac{N}{l}IA\\\end{array}$

Clearly, from the formula of the magnetic flux, the magnetic flux through each turn depends on the current.

Ans: Part a

The magnetic field inside the solenoid is $1.72 \times {10^{ - 4}}{\rm{ T}}$ .

Part b

The magnetic flux through each turn is $4.57 \times {10^{ - 8}}{\rm{ T}} \cdot {{\rm{m}}^2}$ .

Part c

The inductance of the solenoid is $0.54{\rm{ mH}}$ .

Part d.a

The magnetic field inside the solenoid depends on the current.

Part d.b

The magnetic flux through each turn depends on the current.