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A magnet performs 15 oscillations per mi...

A magnet performs 15 oscillations per minute in a horizontal plane, where angle of dip is `60^(@)` and earth is total field is 0.5G. At another place, where total field is 0.6G, the magnet performs 20 Oscillation per minutes. What is the angle of dip at this place.

Text Solution

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As, H=B cos `theta`
Where, H=horizontal component of earth's magnetic field.
B=total eath's mageti field
`and theta="angle of dip"`
`Then, H_(1)=B_(1)cos theta_(1) and H_(2)=B_(2) cos theta_(2)`
`Further, T_(1)=2pi sqrt((I)/(MH_(1)))=2pi sqrt((T)/(MB_(1)cos theta_(1)))`
`and , T_(2)=2pi sqrt((I)/(MH_(2)))=2pi sqrt((T)/(MB_(2)cos theta_(2)))`
` therefore " " (T_(1)^(2))/(T_(2)^(2))=(B_(2)cos theta_(2))/(B_(1)cos theta_(1)`
` therefore " " Cos theta_(2)=(B_(1))/(B_(2))xx (T_(1)^(2))/(T_(2)^(2))cos theta_(1)`
` therefore " " =(B_(1))/(B_(2))xx ((V_(1))/(V_(2)))^(2)cos theta_(1`
` " " (B_(1))/(B_(2))xx((V_(2))/(V_(1)))^(2) cos 60^(@)=0.74`
`therefore " " theta_(2)=cos^(-1)(0.74)=42.2^(@)`
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