Magnetics 5 Question 1
1. A magnetic compass needle oscillates 30 times per minute at a place, where the dip is $45^{\circ}$ and 40 times per minute, where the dip is $30^{\circ}$. If $B _1$ and $B _2$ are respectively, the total magnetic field due to the earth at the two places, then the ratio $\frac{B _1}{B _2}$ is best given by
(2019 Main, 12 April I)
(a) 1.8
(b) 0.7
(c) 3.6
(d) 2.2
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Answer:
Correct Answer: 1. (b)
Solution:
- Given, at first place, angle of dip, $\theta _1=45^{\circ}$
Time period, $T _1=\frac{60}{30}=2 s$
At second place, angle of dip, $\theta _2=30^{\circ}$
Time period, $T _2=\frac{60}{40}=\frac{3}{2} s$
Now, at first place,
$$ B _{H _1}=B _1 \cos \theta _1=B _1 \cos 45^{\circ}=\frac{B _1}{\sqrt{2}} $$
and at second place,
$$ B _{H _2}=B _2 \cos \theta _2=B _2 \cos 30^{\circ}=\frac{\sqrt{3}}{2} B _2 $$
Also, time period of a magnetic needle is given by
$$ \begin{aligned} T & =2 \pi \sqrt{\frac{I}{M B _H}} \\ \therefore \quad T & \propto \sqrt{\frac{1}{B _H}} \text { or } \frac{T _1}{T _2}=\sqrt{\frac{B _{H _2}}{B _{H _1}}} \end{aligned} $$
By putting the values from Eqs. (i) and (ii) into Eq. (iv), we get
$$ \begin{array}{rlrl} \frac{\frac{2}{2}}{2} & =\sqrt{\frac{\sqrt{3} \frac{B _2}{2}}{\frac{B _1}{\sqrt{2}}}} \text { or } \frac{4^{2}}{3}=\frac{\sqrt{3} \times \sqrt{2} B _2}{2 B _1} \\ \Rightarrow \quad & \frac{B _1}{B _2} & =\frac{\sqrt{3} \times \sqrt{2}}{2} \times \frac{9}{16} \\ \Rightarrow \quad & \frac{B _1}{B _2} & =\frac{9 \sqrt{3}}{16 \sqrt{2}} \\ \Rightarrow \quad & \frac{B _1}{B _2} & =\frac{9 \times 1.732}{16 \times 1.414}=\frac{15.588}{22.624} \\ \Rightarrow \quad & \frac{B _1}{B _2} & =0.689 \approx 0.7 T \end{array} $$
