The magnitude J Of the current density In a certain with a circular cross sectio
ID: 1369267 • Letter: T
Question
The magnitude J Of the current density In a certain with a circular cross section of radius R = 3.50 mm is given by J = (5.50 Times 10^8) y^2J , with J in amperes par square mater and radial meters What Is the current through the outer section bounded by r = 0640R and r = R? A human being can be electrocuted If a current as small as 50 mA passes near the heart. An electrician working with sweaty hands makes good contact with the two conductors he is holding if resistance is 2040 Ohm, what might the fatal voltage be ?Explanation / Answer
1)
J = dI/dA
==> dI = J*dA
dI = J*2*pi*r dr
total current form r = 0.64*R to r = R
I = integral dI
= integral J*2*pi*r dr
= integral (5.5*10^8)r^2*2*pi*r dr
= integral 5.5*10^8*2*pi*r^3*dr
= 3.454*10^9 integral r^3*dr
= 3.454*10^9 *(r^4/4) (r value is from 0.64*R tp R)
= (3.454*10^9/4)*( R^4 - (0.64*R)^4)
= (3.454*10^9/4)*(0.8322)*R^4
= (3.454*10^9/4)*(0.8322)*(3.5*10^-3)^4
= 0.108 A
= 108 mA
2)
Apply Ohm's law
v = I*R
= 2040*50*10^-3
= 102 volts
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