A novice electrical engineer mistakenly places two solenoidal inductors near eac

Question

A novice electrical engineer mistakenly places two solenoidal inductors near each other with a common axis on a circuit board. The first coil has a permeability of 2 mT·m/A, a length of 20 mm, a diameter of 10 mm, and 500 turns. The second has exactly half each of these values. Assume for convenience that second catches 10% of the flux from the first, and that the permeability of air is 12.5 T·m/A. Find the mutual inductance in the second coil due to the first. What is the easiest way to fix this mistake so there is almost no mutual inductance at all?

Explanation / Answer

mutual inductance of solenoid 2 wrt 1, M21 = N2*phi21/I1
now, magnetic field due to first solenoid
B = mu*n1*I1
n1 = 500/l1, l1 = 20 mm
n1 = 500/20*10^-3 = 25000
I1 = ?
mu = 2*10^-3
hence
B = 2*10^-3*25000*I1 = 50I1

so phi21 = B*pi*(r1/2)^2 *0.1 = 50I1*pi(5/1000)^2*0.1 = 3.92699*10^-4 I1
hence
M21 = N2*phi21/I1 = 3.92699*10^-4*250 = 0.0981747 H

to rectify this mistake, the solenoid can be taken off acis of the other solenoid and kept side by side where magnetic field due to the first solenoid is almost 0 and hence no mutial inductance

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