Question

A deflection magnetometer is arranged in $\tan A$ position. When a bar magnet is placed deflection is observed to be '$\theta$' and period of oscillation of the needle is ‘$X$’. When the magnet is removed and magnetic needle is made to oscillate in presence of earth’s magnetic field then its period is ‘$Y$’, then $\frac{X}{Y}$ is

• A. $\tan \theta$
• B. $\sqrt{\cos \theta }$
• C. $\sqrt{\sin \theta }$
• D. $\cos \theta$

Answer 1

Answer: (B)

$\sqrt{\cos \theta }$

$\tau =MB\theta =I\alpha =>\alpha ={\omega }^2\theta$

$\omega \propto \sqrt{B_{res}}$

$X=k/\sqrt{B_{res}},Y=k/\sqrt{B_H}$

$X/Y=\sqrt{B_H/B_{res}}$

$B_{res}=B_{H}sec\theta (sinceBr=\sqrt{B_H^2+B^2})$

$X/Y=\sqrt{cos\theta }$