Calculating Self and Mutual Inductance: A Derivation of Formulas Using Solenoids

[toothpaste666]

Homework Statement

Homework Equations

[/B]

∫BdL = μi

Φ = BA (flux)

S= NΦ/i (self inductance)

M = NΦ/i (mutual inductance)

The Attempt at a Solution

a) ∫BdL = μi
BL = μi
B = μi/L
since the solenoid has N1 turns
B = μiN1/L

b) flux Φ = BA = AμiN1/L
self inductance S = N1Φ/i = N1(AμiN1/L)/i = N1^2μA/L

c) mutual inductance M = N2Φ/i = N1N2μA/L

 

[ehild]

Homework Statement

View attachment 83108

Homework Equations

[/B]

∫BdL = μi

Φ = BA (flux)

S= NΦ/i (self inductance)

M = NΦ/i (mutual inductance)

The Attempt at a Solution

a) ∫BdL = μi
BL = μi
B = μi/L
since the solenoid has N1 turns
B = μiN1/L

If i is the current flowing in the wire of the coil, the formula ∫BdL = μi is not correct. You need to show or explain the integration path .

 

[STEMucator]

For part a), I would assume $\vec B$ is uniform within the solenoid and zero outside of it. If you place a rectangular Amperian loop $\partial \Sigma$ with corners $a, b, c, d$ along a cross section of this ideal solenoid, the left side of Ampere's law will give:

$$\oint_{\partial \Sigma} \vec B \cdot d \vec S = \int_a^b \vec B \cdot d \vec S + \int_b^c \vec B \cdot d \vec S + \int_c^d \vec B \cdot d \vec S + \int_d^a \vec B \cdot d \vec S$$

The first integral yields:

$$\int_a^b \vec B \cdot d \vec S = B \int_a^b dS = BL$$

Where $B = |\vec B|$ and $L$ is the length of $\partial \Sigma$ from $a$ to $b$.

The third integral is taken along a segment of $\partial \Sigma$, which lies outside of the solenoid. The magnetic field outside the solenoid is zero, hence the integral is zero.

The second and fourth integrals work out to zero because $\vec B$ is either orthogonal to $d \vec S$, or is equal to zero outside of the solenoid.

Hence we can write:

$$\oint_{\partial \Sigma} \vec B \cdot d \vec S = BL$$

Now using the right side of Ampere's law:

$$BL = \mu_0 i_{enc}$$

The loop $\partial \Sigma$ encloses $nL$ turns where $n$ is the number of turns per unit length of the solenoid. Using this, the enclosed current can be re-written as:

$$i_{enc} = inL$$

Where $i$ is the actual current in the solenoid windings. Hence we can write:

$$BL = \mu_0 inL$$
$$B = \mu_0 in$$
$$B = \mu_0 iN_1$$

This is the magnetic field inside the solenoid.

 

[rude man]

$$B = \mu_0 iN_1/L$$

This is the magnetic field inside the solenoid.

 

[toothpaste666]

Thank you that derivation helped alot. I knew the formula from my book but the derivation of it confused me.

so its
BL = μienc
since ienc = iN1
so
BL = μiN1
B =μiN1/L

Are parts b) and c) ok? I am pretty sure I did part b) right but I am not as confident about part c)

 

[rude man]

Thank you that derivation helped alot. I knew the formula from my book but the derivation of it confused me.

so its
BL = μienc
since ienc = iN1
so
BL = μiN1
B =μiN1/L

Are parts b) and c) ok? I am pretty sure I did part b) right but I am not as confident about part c)

Right on all counts incl. parts b and c!

 

[toothpaste666]

Thank you all

 

[STEMucator]

I always preferred the turns/length version of the derivation, but there is little difference in the final formulas.

Everything else looked good otherwise.