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

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:

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:

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

b) to find the emf you can use:

the emf induced is 14mV

answer: the direction of the wind is expressed as the direction from which the wind is blowing. for example, easterly winds blow from east to west, while westerly winds blow from west to east.

explanation:

the answer would be c