Question

A magnet performs $15$ oscillations per minute in a horizontal plane, where angle of dip is ${60}^{\circ }$ and earth total field is $0.5\text{G}$. At another place, where total field is $0.6\text{G}$, the magnet performs $20$ oscillation per minutes. What is the angle of dip at this place?
$\left[{\cos }^{-1}\right(0.74)={42.2}^{\circ }]$

• A.
  ${42.2}^{\circ }$
• B. ${84.4}^{\circ }$
• C. ${21.1}^{\circ }$
• D. ${11.05}^{\circ }$

Answer 1

The correct option is A

${42.2}^{\circ }$
As the time period of the oscillation of a magnet is,$$T=2\pi \sqrt{\frac{I}{M{B}_H}}$$$\Rightarrow T\propto \frac{1}{\sqrt{B_H}}...\left(1\right)$

Where, $B_H$ is the horizontal component of the earth's magnetic field and $M$ is the magnetic moment of the Earth.

When angle of dip is $\delta$, then horizontal component of earth's magnetic field is,

$B_H=B\cos \delta$

Using, equation $\left(1\right)$ we can write,

$\frac{T_1}{T_2}=\sqrt{\frac{B_{H2}}{B_{H1}}}$

$\frac{T_2^2}{T_1^2}=\frac{B_1\cos {60}^{\circ }}{B_2\cos \delta }......\left(2\right)$

Given, frequency of oscillations,

$f_1=15{\text{min}}^{-1}$ and $f_2=20{\text{min}}^{-1}$

As, $T=\frac{1}{f}$, using equation $\left(2\right)$ we get,

$\frac{{15}^2}{{20}^2}=\frac{0.5\times \frac{1}{2}}{0.6\cos \delta }$

On solving we get,

$\cos \delta =0.74$

$\Rightarrow \delta ={\cos }^{-1}\left(0.74\right)={42.2}^{\circ }$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.