A short bar magnet of magnetic moment the earth's horizontal magnetic field. The distance
where the resultant field makes an angle 60° with from the centre of magnet, at its equatorial position earth's horizontal field is (BH = 0.4 x 10-4 T)
Answers
According to question, the resultant field is inclined at 45
o
with earth's magnetic field.
tanθ=
B
B
H
θ=45
o
tanθ=1=
B
B
H
B
H
=B=
4πr
3
μ
o
M
Given, B= 0.42×10
−4
T
M=5.25×10
−2
J/T
Therefore,0.42+10
−4
=10
−7
×
r
3
5.25×10
−2
r
3
=12.5×10
−5
=125×10
−6
r=5×10
−2
m = 5 cm
(b) At axis of magnet, tan45°=1=
B
H
B
B
H
=B=
4πr
3
μ
o
2M
=>0.42×10
−4
=10
−7
×
r
3
2×5.25×10
−2
=>r
3
=25×10
−5
m
=>r=6.3cm
solution
Explanation:
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12th
Physics
Magnetism and Matter
Earth's Magnetism
A short bar magnet of magne...
PHYSICS
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Asked on December 26, 2019 by
Sharmila Malik
A short bar magnet of magnetic moment 5.25×10
−2
JT
−1
is placed with its axis perpendicular to the earths field direction. At what distance from the centre of the magnet, the resultant field is inclined at 45
∘
with earths field on
(a) its normal bisector and (b) its axis. Magnitude of the earths field at the place is given to be 0.42G. Ignore the length of the magnet in comparison to the distances involved
HARD
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ANSWER
(a) According to question, the resultant field is inclined at 45
o
with earth's magnetic field.
tanθ=
B
B
H
θ=45
o
tanθ=1=
B
B
H
B
H
=B=
4πr
3
μ
o
M
Given, B= 0.42×10
−4
T
M=5.25×10
−2
J/T
Therefore,0.42+10
−4
=10
−7
×
r
3
5.25×10
−2
r
3
=12.5×10
−5
=125×10
−6
r=5×10
−2
m = 5 cm
(b) At axis of magnet, tan45°=1=
B
H
B
B
H
=B=
4πr
3
μ
o
2M
=>0.42×10
−4
=10
−7
×
r
3
2×5.25×10
−2
=>r
3
=25×10
−5
m
=>r=6.3cm
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