Finding the inductance of a coil

[fishingspree2]

Homework Statement

A circular coil has a radius 0.10 m and 30 turns. An external magnetic field 2.60 mT is perpendicular to the coil. a) If there is no current, find the magnetic flux in the coil.
b) When current = 3.8 A in some direction, there is no more net flux in the coil. Find the inductance.

The Attempt at a Solution

I can do a) but I don't understand b. the correct answer of b) is 0.645 mH

a) flux = NB*PI*r^2 = 2.45 miliweber, this is the correct answer

b) L = N*flux / current. If I set flux=0 then I get inductance = 0, which makes no sense because inductance only depends of the geometry.

Please help =(

 

[ideasrule]

Suppose the external magnetic field didn't exist. What would be the magnetic field created by the solenoid? Net field is 0 when this is equal to 2.60 mT.

 

[fishingspree2]

I set B = -2.60 mT

which means that L = N*flux / current = n*-2.60 mT / current?

= (30*2.60*10^-3)/3.8 = 0.0205 H

not correct =(

 

[ideasrule]

The magnetic field produced by the solenoid is B=μ0*N*I/L. When B=2.6 mT, net magnetic field is 0, so B=2.6 mT when I=3.8 A.

Now, what would be the equation for the flux through the solenoid, created by the solenoid itself? You can then find L using L = N*flux / current.

 

[fishingspree2]

Ok, B = μ0*n*i

the length of one turn is 2*Pi*R, the total length of the coil is N times that, so 2*Pi*R*N
n = number of turns / total length = N /[ (2*Pi*R*N) = 1 / (2*Pi*R)

so that B = μ0*i / 2*Pi*R

flux = B*Area = B*Pi*(R^2) = μ0*i*r / 2
so that L = N*flux / i = μ0*r*N / 2 = 1.88 * 10^-6 W, wrong answer

I have also tried multiplying the flux by N since there are N turns

please help =(

 

[Fazza3_uae]

Ideasrule told you that ...


B=2.6 mT when I=3.8 A ,,,,,, So find L the length of the solenoid by this equstion :

B=μ₀*N*I/L

Where ,

B = 2.6e^-3 T
N = 30 turns
I = 3.8 A
So L = ?

Then substitute the value of L in this equation to find the Inductance of the Solenoid :

L _{Solenoid's Inductance} = μ₀*N²*A / L _{Length Found}

Thats it .


Just for more clarification here are the steps to get the last equation for solenoid's inductance :

B _Solenoid = μ₀*n*I

n = N/L , So B _Solenoid = μ₀*N*I/L

------------------------------------------------------------

\phi _B = N*B*A*Cos(0⁰)

B _Solenoid = μ₀*N*I/L

So , \phi _B = N*(μ₀*N*I/L )*A*Cos(0⁰)

Cos(0⁰) = 1

So , \phi _B = μ₀*N²*I*A / L


-------------------------------------------------------------------------

The emf induced in N turns is N times the emf in one turn :

\epsilon = - d\phi _B / dt

\epsilon = - ( μ₀*N²*A / L ) * (dI/dt)

--------------------------------------------------------------------------

Finally the inductance :

L _{Solenoid's Inductance} = - \epsilon / (dI/dt)

We will get finally :

L _{Solenoid's Inductance} = μ₀*N²*A / L _{Length Found}

Good Luck ...