Calculating the induced emf in a solenoid, have worked on it, can't get answer

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SUMMARY

The discussion focuses on calculating the induced electromotive force (emf) in a solenoid with specific parameters: a length of 3.7 cm, a diameter of 0.95 cm, and 160 turns per cm. The rate of current decrease is 38.0 A/s. The formula used for the induced emf is emf = -dΦ/dt = μ0N²(A/L) di/dt, where μ0 is the permeability of free space (4π × 10-7 H/m). The user initially calculated the induced emf for one winding incorrectly and later attempted to calculate it for the entire solenoid using N=160, resulting in an incorrect value of 234 milliamps.

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Homework Statement


A solenoid of length 3.7 cm and diameter 0.95 cm is wound with 160 turns per cm. The current is decreasing at a rate of 38.0 A/s.
(a) What is the induced emf in one of the windings?

(b) What is the induced emf in the entire solenoid?


Homework Equations




emf = -dΦ/dt = μ0N²(A/L) di/dt
Given:
μ0 is the permeability of free space = 4π × 10-7 H/m
N is the number of turns
A is the area where the magnetic flux passes through = π(d/2)²
d is the diameter of the solenoid
L is the length of the solenoid
di/dt is the rate of change in the current
For the first part I set N=1 and for the second I set N=160

The Attempt at a Solution


4pi * 10^-7 (1)^2 (pi(.95)/2)^2/3.7 (36)
I get 9.15 E-6 which is not the correct answer and for B when I plug in the second set of numbers I get 234 milliamps or .234 amps
 
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What is B for a solenoid? Note that the expression is in terms of the turns per length ... do not simply use N=1 to calculate B.
 

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