15. A short bar magnet of magnetic moment
5v3 x 10-2 JT is placed with its axis normal to
the earth's horizontal magnetic field. The distance
from the centre of magnet, at its equatorial position,
where the resultant field makes an angle 60° with
earth's horizontal field is (BH = 0.4 * 10-4 T)
[NCERT Pg. 180]
1
(1) 5(3) cm
(2) 5 cm
(3) 5(6) cm
(4) 10 cm
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Answer is (a)
Explanation:
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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