Question

In: Physics

A 36.5 mA current is carried by a uniformly wound air-core solenoid with 430 turns, a...

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

Solutions

Expert Solution

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=μ0NliB = {\mu _0}\frac{N}{l}i

Here, μ0{\mu _0} is the permeability of vacuum, NN is the number of turns of the solenoid, ll is the length of the solenoid and ii is the current through the solenoid.

The magnetic flux through the coil is given as follows:

Φ=BA\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=πr2A = \pi {r^2}

Here, rr is the radius of solenoid.

The self-inductance of the solenoid is given as,

L=NΦIL = \frac{{N\Phi }}{I}

Here, μ0{\mu _0} is the permeability of vacuum, NN is the number of turns of the solenoid, AA is the area of cross section of the solenoid, and ll is the length of the solenoid.

(a)

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

Substitute 4π×107Tm/A4\pi \times {10^{ - 7}}{\rm{ T}} \cdot {\rm{m/A}} for μ0{\mu _0} , 430 for N, 11.5 cm for ll , and 36.5mA36.5{\rm{ mA}} for ii in the equation B=μ0NliB = {\mu _0}\frac{N}{l}i and calculate the magnetic field inside the solenoid.

B=(4π×107Tm/A)(430)(11.5cm)(36.5mA)=(4π×107Tm/A)(430)(11.5cm(102m1cm))(36.5mA(103A1mA))=1.72×104T\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.

Φ=BA\Phi = BA

Substitute πr2\pi {r^2} for AA and d2\frac{d}{2} for rr in the equation Φ=BA\Phi = BA .

Φ=B(πr2)=B(π(d2)2)=Bπd24\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 1.72×104T1.72 \times {10^{ - 4}}{\rm{ T}} for BB , and 18.5mm18.5{\rm{ mm}} for dd in the equation Φ=Bπd24\Phi = B\pi \frac{{{d^2}}}{4} .

Φ=(1.72×104T)π(18.5mm)24=(1.72×104T)π(18.5mm(103m1mm))24=4.57×108Tm2\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, 4.6×108Tm24.6 \times {10^{ - 8}}\,{\rm{T}} \cdot {{\rm{m}}^2} for Φ\Phi , and 36.5 mA for ll in the equation L=NΦIL = \frac{{N\Phi }}{I} and calculate L.

L=(430)(4.6×108Tm2)(36.5mA)=(430)(4.6×108Tm2)(36.5mA(103A1mA))=5.42×104H(103mH1H)=0.54mH\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 μ0NlI{\mu _0}\frac{N}{l}I for BB in the equation of magnetic flux Φ=BA\Phi = BA .

Φ=BA=μ0NlIA\begin{array}{c}\\\Phi = BA\\\\ = {\mu _0}\frac{N}{l}IA\\\end{array}

Substitute μ0NlIA{\mu _0}\frac{N}{l}IA in the equation of self-inductance L=NΦIL = \frac{{N\Phi }}{I} .

L=N(μ0NlIA)I=N2μ0Al\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=μ0NlIB = {\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 μ0NlI{\mu _0}\frac{N}{l}I for BB in the equation of magnetic flux Φ=BA\Phi = BA .

Φ=BA=μ0NlIA\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 1.72×104T1.72 \times {10^{ - 4}}{\rm{ T}} .

Part b

The magnetic flux through each turn is 4.57×108Tm24.57 \times {10^{ - 8}}{\rm{ T}} \cdot {{\rm{m}}^2} .

Part c

The inductance of the solenoid is 0.54mH0.54{\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.


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