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.

Question

ROSHAN MUJEEB · Answer

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

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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.

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.

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