Question

. Two magnetic poles of strengths 'm¹' and 'm²'
when placed in air at a distance of d from each
other, a force 'F' acts between them. If the pole
strength of the poles is doubled, the distance
between them is trebbled and the poles are placed
in some other medium, the force between them
reduces by 25%. Find the relative magnetic per
meability of the medium.​

Answer 1

as apply two conditions and then solve it.

Answer 2

Answer:

Explanation:

1st case :

let the magnetic pole strength1 be "m1"

let the magnetic pole strength2 be "m2"

let the distance between the poles be "r1"

therefore the  magnetic force of attraction working between them will be:

F1 = (μo*m1*m2/4π*r²)      [  F ∝ m1 * m2  && F ∝ r² thus (μ0*m1*m2)/(4π*r²)]

case 2:

as per the question pole strength1 now becomes : 2*m1

similarly pole strength2 becomes: 2*m2

the distance between the poles becomes: 3*r1

hence magnetic force2 becomes:

F2 = (μ * 4*m1*m2)/(4π*9r²)

According to the question :

F1 = F2*(100-25)%

Therefore :

(μo*m1*m2/4π*r²) = [(μ * 4*m1*m2)/(4π*9r²) ] * 75/100

or (μo)=(μ*4/9) * 3*25/100

or  μo=(μ*4/9) *3/4

or  μo = μ/3

or μ/μo = 3

but μ/μo is given as μr

therefore μr = 3 .

Answer: THE RELATIVE PERMEABILITY OF THE MEDIUM IS 3.