What is the induced emf in one of the windings?

[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
radius r = d/2
= 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 \

 

[TSny]

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

 

[lizaliiu]

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

 

[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.

 

[Nickie]

The induced emf in one of the windings can be calculated using Faraday's Law of Induction, which states that the induced emf is equal to the negative of the rate of change of magnetic flux through the coil.

In this case, the magnetic flux through the coil is changing due to the decreasing current. The magnetic flux is given by the product of the number of turns in the coil, the area of the coil, and the magnetic field. Therefore, we can calculate the induced emf as follows:

ε = -NΔΦ/Δt

where N is the number of turns, ΔΦ is the change in magnetic flux, and Δt is the change in time.

First, we need to calculate the change in magnetic flux. The magnetic field inside a solenoid is given by B = μ0nI, where μ0 is the permeability of free space, n is the number of turns per unit length, and I is the current.

Using this equation, we can calculate the initial magnetic field inside the solenoid when the current is 36.3 A/s:

B = μ0nI = (4π*10^-7 T.m/A)(170 turns/cm)(36.3 A/s) = 2.47*10^-3 T

The change in magnetic flux is then given by ΔΦ = BA = (2.47*10^-3 T)(6.55*10^-5 m^2) = 1.62*10^-7 Wb

Plugging these values into the equation for induced emf, we get:

ε = -(1 turn)(1.62*10^-7 Wb)/(1 s) = -1.62*10^-7 V

Therefore, the induced emf in one of the windings is -1.62*10^-7 V.

For the entire solenoid, we simply need to multiply this value by the total number of turns in the coil:

ε = (-1.62*10^-7 V)(170 turns/cm)(3.64 cm) = -9.01*10^-5 V

The negative sign indicates that the induced emf is in the opposite direction of the decreasing current.