# Homework Help: Finding the inductance of a coil

1. Dec 20, 2009

### fishingspree2

1. The problem statement, all variables and given/known data
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.

3. 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.

Last edited: Dec 20, 2009
2. Dec 20, 2009

### 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.

3. Dec 20, 2009

### 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 =(

4. Dec 20, 2009

### 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.

5. Dec 20, 2009

### 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

6. Dec 20, 2009

### Fazza3_uae

Ideasrule told you that ...

B=2.6 mT when I=3.8 A ,,,,,, So find L the lenght 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 ...

Last edited: Dec 20, 2009