First slide

Magnetism

Question

A magnet suspended so as to swing horizontally makes 50 vibrations per minutes at a place where dip is $30^{\circ}$ , and 40 vibrations where dip is $45^{\circ}$ . Compare the earth's total fields at the two places.

Moderate

A

$\frac{25}{8 \sqrt{6}}$
B

$\frac{50}{8 \sqrt{6}}$
C

$\frac{25}{4 \sqrt{6}}$
D

$\frac{50}{4 \sqrt{6}}$

Solution

We know time period of oscillation of magnetic needle
$T = 2 π \sqrt{\frac{I}{M B_{H}}} (∵ Frequency =n= \frac{1}{T})$
$\Rightarrow n \propto \sqrt{B_{H}}$
$\Rightarrow \frac{n_{1}}{n_{2}} = \sqrt{\frac{B_{H_{1}} \cos θ_{1}}{B_{H_{2}} \cos θ_{2}}} [T a k e H o r i z o n t a l c o m p o n e n t o f f i e l d]$
Given data $n_{1} = 50, n_{2} = 40, θ_{1} = 30 °, θ_{1} = 45 °$
$\Rightarrow \begin{array}{l} \Rightarrow \frac{50}{40} = \sqrt{\frac{B_{H_{1}}}{B_{H_{2}}}} \times \sqrt{\frac{\cos 30 °}{\cos 45 °}} \\ \frac{25}{16} = \frac{B_{H_{1}}}{B_{H_{2}}} \frac{\sqrt{3}}{\sqrt{2}} \\ \Rightarrow \frac{B_{H_{1}}}{B_{H_{2}}} = \frac{25}{8 \sqrt{6}} \end{array}$

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