2069 Q.No. 27 A bar magnet is placed in the magnetic meridian with its North pole pointing
geographical north. If the neutral points are observed at 18 cm from each pole. Find the
magnetic moment of the magnet? [H = 3.4 x 10-4 T]
Ans: 2.01 Am(please solve it it's urgent)
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12th
Physics
Magnetism and Matter
Properties of Magnetic Field
A bar magnet 30 cm long is ...
PHYSICS
A bar magnet 30 cm long is placed in the magnetic meridian with its north pole pointing south. The neutral point is observed at a distance of 30 cm from its one end. Calculate the pole strength of the magnet. Given horizontal component of earth's field =0.34 G.
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ANSWER
Here, 2l=30cm
or l=15cm=0.15m,
t=30cm=0.30cm,
B
H
=0.34G=0.34×10
−4
T
When magnet is placed with its north pole pointing south, neutral point is obtained on its axial line. Therefore, at the neutral point.
B
axial
=B
H
or
4π
μ
0
×
(r
2
−l
2
)
2
2Mr
=B
H
or M=
μ
0
4π
×
2r
B
H
(r
2
−l
2
)
2
=
10
−7
1
×
2×0.30
0.34×10
−4
×(0.30
2
−0.15
2
)
2
=
10
−7
×2×0.30
0.34×10
−4
×(0.0675)
2
=2.582Am
2
The pole strength of the magnet,
m=
2l
M
=
0.30
2.582
=8.606Am
hope is this helpful for you
Answer:
1.98Am^2
Explanation:
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