Question

(a) Calculate the self-inductance of a solenoid that is

Answer 1

a) 0.142mH

b) 14mV

Step-by-step explanation:

the complete answer is:

(a) Calculate the self-inductance of a solenoid that is <ghtly wound with wire of diameter 0.10 cm, has a cross-sec<onal area of 0.90 cm2 , and is 40 cm long. (b) If the current through the solenoid decreases uniformly from 10 to 0 A in 0.10 s, what is the emf induced between the ends of the solenoid

a) the self inductance of a solenoid is given by:

`L=(\mu_o N^2 A)/(L)`

μo: magnetic permeability of vacuum = 4\pi*10^{-7}N/A^2

A: cross sectional area = 0.9cm^2=9*10^{-5}m

L: length of the solenoid = 40cm = 0.4m

The N turns of the wire is calculated by using the diameter of the wire:

N = (40cm)/(0.10cm)=400

By replacing in the formula you obtain:

`L=((4\pi *10^(-7)N/A^2)(400)^2(9*10^(-5)m^2)))/((0.4m))=1.42*10^(-4)H`

the self inductance is 1.42*10^{-4}H = 0.142mH

b) to find the emf you can use:

`emf=L(\Delta I)/(\Delta t)=(1.42*10^(-4)H)(10A-0A)/(0.10s)=0.014V=14mV`

the emf induced is 14mV