Question

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​

Answer 1

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