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)`.

Explanation

(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")` )