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a deflection magnetometer is adjusted in the usual way.when a magnetometer is introduced,the deflection observed is @(theta),and the period of oscillation of the needle in the mangetometer is T.when the magnet is removed the period of oscillations is T0.the relation between Tand T0 is A)T^2=T0^2 Cos @


Navjyot Kalra , 11 Years ago
Grade 10
anser 1 Answers
ROSHAN MUJEEB

Last Activity: 4 Years ago

When deflection magnetometer is rest in usual way the field due to magnet (F) and horizontal component (H) of earth's field are perpendicular to each other. In this setting
T=2πIMF2+H2−−−−−−−√−−−−−−−−−−−−√...(i)T=2πIMF2+H2...(i)
After removing the magnet
T0=2πIMH−−−−−√...(ii)T0=2πIMH...(ii)
We know,FH=tanθFH=tanθ
Dividing (i) by (ii), we get,
TT0=HF−−√2+H2−−−−−−−−−−√TT0=HF2+H2
=HH2tan2θ+H2−−−−−−−−−−−−√−−−−−−−−−−−−−−−√=HHsec2θ−−−−−√−−−−−−−−√=cosθ−−−−√=HH2tan2θ+H2=HHsec2θ=cosθ
⇒T2T20=cosθ∴T2=T20cosθ

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