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A short magnet oscillation in vibration ...

A short magnet oscillation in vibration magnetometer with a frequency 10Hz. A downward current of 15A is established in a long vertical wire placed 20cm to the West of the magnet. The new frequency of the short magnet is (the horizontal of the component of earth's magnetic field is `12mu`)

A

4Hz

B

2.5Hz

C

9Hz

D

15Hz

Text Solution

Verified by Experts

The correct Answer is:
D
Frequency, `v=(1)/(2pi) sqrt(MH)/(I)`
`v=(1)/(2pi) sqrt(M(B+H))/(I)`
where, B=magnetic field due to downward conductor

`Then " " B=(mu_(0))/(4pi).(2I)/(a)`
` " " B=10^(-7)xx(2xx15)/(20xx10^(-2))`
`B=15muT`
` " " (v')/(v)=sqrt((B+B_(H))/(B))`
` " " (v')/(10)=sqrt((15+12)/(12))=1.5`` " " v'=15Hz`
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