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A solenoid has a core of material with relative permeability 400. The winding of the solenoid are insulted from the core and carry a current of 2A. If the number of turns is 1000 per metre, calculate (a) H, (b) M, (c) B and (d) the magnetising current i_(m). |
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Answer» Solution :(a) The FIELD H is dependent of the material of the core, and is `H=nI=100 xx 2.0=2 xx 10^(3) A//m` (b) The magnetic field B is given by `B=mu_(r) mu_(0)H` `=400 xx 4pixx10^(-7) (N//A^(2)) xx 2 xx 10^(3) (A//m)` =1.0T (c) Magnetisation is given by `M=(B-mu_(0) H)//mu_(0)` `=(mu_(r) mu_(0) H-mu_(0) H)=(mu_(r)-1)=399 xx H` `=8 xx 10^(5) A//m` (d) The magnetising current `I_(M)` is the additional current that needs to be passed through the WINDING of the solenoid in the absence of the core which would give a B value as in the PRESENCE of the core. Thus `B=mu_(r) n(I+I_(M))`. Using I=2A, B=1 T, we get `I_(M)=494A`. |
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