A magnetic pole of bar magnet with pole
strength of 100 A m is 20 cm away from
the centre of a bar magnet. Bar magnet has
pole strength of 200 A m and has a length
5 cm. If the magnetic pole is on the axis
of the bar magnet, find the force on the
magnetic pole.
Answers
Answered by
4
Answer:
#BAL Here, 2l=10cm, l=5cm=0⋅05m.
m=10Am
B1=?,B2=?,d=0⋅2m,
M=m×2l=10×0⋅1=1Am2
B1=Baξal=μ04π2Md(d2-l2)2=10-7×2×1×0⋅2(0⋅22-0⋅052)2
=0⋅4×10-7(0⋅0375)2=2⋅84×10-5T
B2=Bequatorial
=μ04πM(d2+l2)3/2=10-7×1(0⋅22+0⋅052)3/2
=10-7(0⋅0425)3/2=10-78⋅76×10-3=1⋅14×10-5T
Explanation:
Answered by
0
Thus the force on the magnet pole is F = 500 x 10^3 N
Explanation:
Given data:
- Pole strength of the bar magnet = 100 Am
- Pole strength of another bar magnet = 200 Am
- Distance between the the two bar magnets = 20 cm
Solution:
We know that the force between the bar magnets are,
F = μ q1 q2 / 4 π r^2
F = 4 x π x 100 x 200 / 4 π x 0.20^2
F = 20000 / 0.04
F = 500 x 10^3 N
Thus the force on the magnet pole is F = 500 x 10^3 N
Similar questions