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A solenoid has a core of a material with relative permeability 400. The windings of the solenoid are insulated 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) Magnetising intensity (independent of material of core) is given by `H= i_f` (Where `i_f=` magnetising current per unit length of solenoid) `H=nI_f` (Where `I_f=` magnetising current, `N=` no. of turns per unit length of solenoid) `THEREFORE H = (1000) (2)` `therefore H= 2 xx 10^(3) Am^(-1)` (b) Total magnetic field in a given magnetic core, `B= mu_(I) mu_(0) H` `= (400) (4pi xx 10^(-7) ) (2 xx 10^(3) )` `therefore B= 1.0048 T` (c) Magnetisation obtained in a given magnetic core, `M= (B)/( mu_0) - H ( because H= (B)/( mu_0) - M)` `= (1.0048)/( 4pi xx 10^(-7) ) -2 xx 10^(3)` `therefore M=7.98 xx 10^(5) Am^(-1)` (NOTE : Here `M=i_b=` INDUCED current (or bound current) per unit length of magnetic core) (d) Additional magnetising current is given by, `I_(m) = (M)/( n)` `= (7.98 xx 10^(5) ) /( 1000)` `therefore I_(m) = 798 A` (Note : Unit of H is actually `("(ampere) (turn)")/( "(metre)")` and unit of `I_m` is actually `("ampere")/("turn")` ) |
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