Question

A deflection magnetometer is adjusted and a magnet of magnetic moment $M$ is placed on it in the usual manner and the observed deflection is $\theta$. The period of oscillation of the needle before settling to the deflection is $T$. When the magnet is removed, the period of oscillation of the needle is $T_0$ before settling to $0^o$. If the earth's induced magnetic field is $B_H$, the relation between T and $T_0$ is :

• A. $T^2=T_0^{2}cos\theta$
• B. $T^2=\frac{T_0^2}{cos\theta }$
• C. $T=T_{0}cos\theta$
• D. $T=\frac{T_0}{cos\theta }$

Answer 1

The correct option is A $T^2=T_0^{2}cos\theta$

Let field due to magnet be $B$ and due to earth's magnetic field be $B_H$.

By Tangent Law, $B=B_H\tan \left(\theta \right)$

In deflection magnetometer, net field is

$B_{net}=\sqrt{B^2+B_H^2}$

$=\sqrt{B_H^2+B_H^{2}ta{n}^2\left(\theta \right)}$

$=\frac{B_H}{cos\theta }$

Time period with magnet is given by,

$T=2\pi \sqrt{\frac{I}{M{B}_{net}}}$

Time period when magnet is removed,

$T_o=2\pi \sqrt{\frac{I}{M{B}_H}}$

Hence, $\frac{T}{T_o}=\sqrt{\frac{B_H}{B_{net}}}$

$⟹T^2=T_o^2\cos \theta$