A bar magnet has pole strength `4*5Am`, magnetic length `12cm` and cross sectional area `0*9cm^2`. Find (a) intensity of magnetisation (I) (b) magnetising intesity (H) at the centre and (c) magnetic induction (B) at the centre of the magnet.

Answer

`5xx10^4Am^-1`, `199Am^-1`, `6*26xx10^-2T`

Explanation

Here, `m=4*5Am.`,
`2l=12cm=0*12m`
`a=0*9cm^2=0*9xx10^-4m^2`
(a) `I=m/a=(4*5)/(0*9xx10^-4)`
`=5xx10^4Am^-1` from S to N
(b) Magnetic intensity at the centre, due to its N-pole
`H_1=(m)/(4pir^2)=(4*5)/(4pi(0*06)^2)` where `r=2l//2`
`=99*5Am^-1`, towards S-pole
Magnetic intensity at the centre due to its S pole
`H_2=(m)/(4pir^2)=(4*5)/(4pi(0*06)^2)`
`99*5Am^-1`, towards S-pole.
Resultant magnetic intensity at the centre
`H=H_1+H_2=99*5+99*5`
`=199Am^-1`, towards S-pole
(c) Now, `B=mu_0(H+I)`
`=4pixx10^-7(-199+5xx10^4)`
`=6*26xx10^-2T`, towards N pole