If F1 and F2 are the strengths of magnetic fields at same distance ‘d' , from the centre of a bar magnet, on axial and equatorial lines respectively, show that F1/F2 =2+7l^2/d^2 , where ‘l’ is the magnetic length.

We know that the magnetic field at a distance d from centre on the axis of a magnet of length 2l and magnetic moment M is given by:-
B=
4π
μ
o
​

​

(d
2
−l
2
)
2

2Md
​

For P
1
​
, d
1
​
=10cm
for P
2
​
, d
2
​
=d
1
​
+10=20cm
B
1
​
=
4π
μ
o
​

​

(d
1
2
​
−l
2
)
2

2Md
1
​

​

B
2
​
=
4π
μ
o
​

​

(d
2
2
​
−l
2
)
2

2Md
2
​

​

Given that :
B
2
​

B
1
​

​
=
2
25
​

⟹
d
2
​

d
1
​

​

(d
1
2
​
−l
2
)
2

(d
2
2
​
−l
2
)
2

​
=
2
25
​

⟹
20
10
​

(10
2
−l
2
)
2

(20
2
−l
2
)
2

​
=
2
25
​

⟹
(10
2
−l
2
)
2

(20
2
−l
2
)
2

​
=25
⟹
10
2
−l
2

20
2
−l
2

​
=5
⟹400−l
2
=500−5l
2

⟹4l
2
=100
⟹l
2
=25
⟹l=5cm
Hence,magnetic length = 2l=10cm