A short magnet oscillates in vibration magnetometer with a frequency 10
where horizontal component of earth's magnetic field is 12 MT. A downw
current of 15A is established in a long vertical wire placed 20 cm west of
magnet. New frequency is
A) 2.5 Hz
B) 5 Hz
C) 9 Hz
D) 4 Hz
Answers
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B
H
=24×10
−6
T,
T
1
=0.1
′
B = B
H
−B
wire
=2.4×10
−6
−
2πr
μ
0
i
= 24×10
−6
−
0.2
2×10
−7
×18
= (24−10)×10
−6
= 14×10
−6
T = 2π
MB
H
I
T
2
T
1
=
B
H
B
T
2
0.1
=
24×10
−6
14×10
−6
(
T
2
0.1
)
2
=
24
14
T
2
2
=
24
(0.01×14)
T
2
=0.0766s
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