# What is the induced emf in one of the windings?

1. Feb 10, 2013

### lizaliiu

1. A solenoid of length 3.64 cm and diameter 0.914 cm is wound with 170 turns per cm. If the current is decreasing at a rate of 36.3 A/s, what is the induced emf in one of the windings?

this is what I did:
Length of the solenoid l = 3.64cm
= (3.64 cm)(10-2 m/ 1 cm)
= 0.0364 m
diameter d = 0.914 cm
= 0.914 cm /2
= 0.457 cm
= (0.457 cm)(10-2 m/ 1 cm)
= 0.00457 m
area A = πr2
= (3.14)(0.00457 m)2
= 6.55*10-5 m2
the change of rate of current dI/dt = 36.3 A/s
_______________________________________
a)
for one winding N = 1
for one winding, n = N/1 cm
= 1/10-2 m
= 100 m
per one meter of winding,
n = 100
the induced emf in one of the windings is
ε = [μ0 n2A/l](dI/dt)
= [(4π*10-7 T.m/A)(100)2(6.55*10-5 m2 )/(0.0364 m)](36.3 A/s)
= 1.64*10-5 V <--------BUT THE ANSER IS WRONG...
B. What is the induced emf in the entire solenoid?

This is what I did:
for entire solenoid, n = 170 turns/ 1 cm
= 170 / 10-2 m
= 17000 m
per one meter, n = 17000
the induced emf in the entire solenoid is
ε = [μ0 n2A/l](dI/dt)
= [(4π*10-7 T.m/A)(17000)2(6.55*10-5 m2 )/(0.0364 m)](36.3 A/s)
but the answer was still wrong....
could someone help me where did i do wrong, please \
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Feb 10, 2013

### TSny

Looks like you are using the formula B = μo(N/L)I for the field inside a solenoid. What do N and L stand for here?

3. Feb 11, 2013

### lizaliiu

N=how many turns
L=length of the wire?!

4. Feb 11, 2013

### TSny

N is the total number of turns and L is the total length of the solenoid. So, N/L is the number of turns per unit length, which you found to be 17000 turns per m.