T2=T20cosθ
T=T0cosθ
T2=T20cosθ
T2=T0cosθ
The correct option is A T2=T20cosθ
In the usual setting of deflection magnetometer, field due to magnet(F)and horizontal component(H)of earth's field are perpendicular to each other. Therefore, the net field on the magnetic needle is √F2+H2
∴T=2π√IM√F2+H2 (i)
When the magnet is removed,T0=2π√IMH (ii)
Also, F/H=tanθ
Dividing (i) by (ii), we get
T/T0=√H√F2+H2=√H√H2tan2θ+H2=√H/Hsecθ=cosθ
→T2/T20=cosθ
∴T2=T20cosθ.