A deflection magnetometer is placed with its arm along the east-west direction and a bar magnet is placed along the arm of the magnetometer. Due to the magnet, 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 T 0 . The relation between T and T 0 is

Question

A deflection magnetometer is placed with its arm along the east-west direction and a bar magnet is placed along the arm of the magnetometer. Due to the magnet, 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 T 0 . The relation between T and T 0 is, T 2 = T 0 2 cos ⁡ θ, T = T 0 cos ⁡ θ, T = T 0 cos ⁡ θ, T 2 = T 0 2 cos ⁡ θ

MARKS App · Accepted Answer

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 F 2 + H 2 ∴ T = 2 π I M F 2 + H 2 ...(i) When the magnet is removed, T 0 = 2 π I MH ...(ii) Also, F H = tan ⁡ θ Dividing (i) by (ii), we get T T 0 = H F 2 + H 2 = H H 2 tan 2 ⁡ θ + H 2 = H H sec 2 ⁡ θ = cos ⁡ θ ⇒ T 2 T 0 2 = cos ⁡ θ ∴ T 2 = T 0 2 cos ⁡ θ

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