A deflection magnetometer is adjusted in the usual way. When a magnet is introduced, the deflection observed is θ, and the period of oscillation of the needle in the magnetometer is T. When the magnet is removed, the period of oscillation is T₀. The relation between T and T₀ is

T² = $T_{0}^{2}$ cosθ

T = T₀ cosθ

T = $\frac{T_{0}}{\cos θ}$

$T^{2} = \frac{T_{0}^{2}}{cosθ}$

Solution

T² = $T_{0}^{2}$ cosθ

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 $\sqrt{F^{2} + H^{2}}$

∴ T = 2 $π \sqrt{\frac{I}{M \sqrt{F^{2} + H^{2}}}}$ ....(i)

When the magnet is removed

T₀ = 2 $π \sqrt{\frac{I}{MH}}$ .....(ii)

Also, $\frac{F}{H}$ = tanθ

Dividing (i) by (ii), we get

$\frac{T}{T_{0}} = \sqrt{\frac{H}{\sqrt{F^{2} + H^{2}}}}$

= $\sqrt{\frac{H}{\sqrt{H^{2} \tan^{2} θ + H^{2}}}}$

= $\sqrt{\frac{H}{H \sqrt{\sec^{2} θ}}}$

= $\sqrt{\cos θ}$

$\frac{T^{2}}{T_{0}^{2}}$ = cosθ

∴ T₀ = $T_{0}^{2}$ cosθ

For Daily Free Study Material Join wiredfaculty

Electrostatic Potential And Capacitance

Some More Questions From Electrostatic Potential and Capacitance Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Loaction