# Stuck on Cylindrical iron-clad solenoid magnet problem

Thread Starter

#### Mang Ge

Joined Oct 14, 2015
3
Guys, Im new here but have been stuck on this question for like 2 days 5 hours till now. Pls help me out of this thing.

a) Draw the magnetic equivalent circuit.
b) Compute the flux density in the working air gap for x=10 mm.
c) Compute the value of the energy stored in Wfld (for x=10 mm).
d) Compute the value of the inductance L (for x=10 mm).
e) For a force ffld of 1000 N determine for x=10 mm the current I=I0 required.

Values given
grad = 1 mm, R=20 mm, and d=10 mm. The exciting coil has N=1000 turns and carries a constant current of I=10A.

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Thread Starter

#### Mang Ge

Joined Oct 14, 2015
3

#### recklessrog

Joined May 23, 2013
985
Sorry but I'm a little confused, looking at your drawing of the core, it shows 1000 turns, but your current is a CONSTANT 10 amperes. This would require (assuming copper wire) close to 15 swg wire (10.5 amps) 26.8 turns sqcm. So there does not appear to be enough winding window room available. Am I missing something?

Thread Starter

#### Mang Ge

Joined Oct 14, 2015
3
Sorry but I'm a little confused, looking at your drawing of the core, it shows 1000 turns, but your current is a CONSTANT 10 amperes. This would require (assuming copper wire) close to 15 swg wire (10.5 amps) 26.8 turns sqcm. So there does not appear to be enough winding window room available. Am I missing something?
Hi, thanks for replying. This is what this problem exactly says. But I didn't have the thought as deep as yours.

The cylindrical iron-clad solenoid magnet shown in Figure 1 has a plunger which can move a relatively short distance x developing a large force ffld. The plunger is guided so that it can move in vertical direction only. The radial air gap between the shell and the plunger is grad = 1 mm, R=20 mm, and d=10 mm. The exciting coil has N=1000 turns and carries a constant current of I= 10A