Question

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

Answer 1

Answer:

a) 0.142mH

b) 14mV

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:

$L=\frac{\mu_o N^2 A}{L}$

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:

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

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

b) to find the emf you can use:

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

the emf induced is 14mV