Question

A magnetic compass needle oscillates $30$ times per minute at a place where the dip is $45°$, and $40$ times per minute where the dip is $30º$. If B1 and B2 are respectively the total magnetic field due to the earth at the two places, then the ratio B₁/ B₂ is best given by:

• A. $1.8$
• B. $2.2$
• C. $0.7$
• D. $3.6$

Answer 1

The correct option is C

$0.7$

Step 1: Given data

For the first magnetic needle frequency$\left(f_1\right)$ = $30mi{n}^{-1}$and ${\theta }_1=45°$

And, for the second magnetic needle frequency $\left(f_2\right)=30mi{n}^{-1}$and ${\theta }_2=30°$

Step 2: Formula used

The following expression describes the relationship between the frequency of motion of a magnetic needle and the magnetic field at a given location:

$f=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mu B_H}{I}}$

Here $\mu$ is the magnetic moment of the magnetic needle, $B_H$ denotes the horizontal component earth’s magnetic field while $I$ represent the moment of inertia of the magnetic needle about the axis about which it oscillates.

Step 3: Relation between frequency and magnetic field

The frequency of oscillations of a magnetic needle in the earth’s magnetic field is given as:
$f=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mu B_H}{I}}$, and $B_H=B\cos \left(\theta \right)$, where $\theta$ is known as the angle of dip which the magnetic field (B) makes with the horizontal axis. Thus,

$\Rightarrow f=\frac{1}{2\mathrm{\pi }}\sqrt{\frac{\mu B\cos \left(\theta \right)}{I}}$

We can observe from the above equation that

$f\propto \sqrt{B\cos \left(\theta \right)}$

Step 4: Expression for both the needles
For the first needle,

$f_1\propto \sqrt{B_1\cos \left({\theta }_1\right)}\dots \left(i\right)$

For the second needle,

$f_2\propto \sqrt{B_2\cos \left({\theta }_2\right)}\dots \left(ii\right)$

Diving equation (i) with equation (ii),

$\frac{f_1}{f_2}=\frac{\sqrt{B_1\cos \left({\theta }_1\right)}}{\sqrt{B_2\cos \left({\theta }_2\right)}}$

Inserting the known values,

$\frac{30}{40}=\frac{\sqrt{B_1\cos \left(45°\right)}}{\sqrt{B_2}\cos \left(30°\right)}$

Squaring both sides,

$$\begin{gathered} \frac{900}{1600}=\frac{B_1\times {\displaystyle \frac{1}{\sqrt{2}}}}{B_2\times {\displaystyle \frac{\sqrt{3}}{2}}} \\ \Rightarrow \frac{9}{16}=\sqrt{\frac{2}{3}}\times \frac{B_1}{B_2} \\ \Rightarrow \frac{B_1}{B_2}=\frac{9}{16}\times \sqrt{\frac{2}{3}} \\ \Rightarrow \frac{B_1}{B_2}=0.68\simeq 0.7 \end{gathered}$$

Hence, the correct answer is option (C).