Magnetics Ques 98
- 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: 98.(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}} \quad …….(i) $
and at second place,
$ B _{H _2}=B _2 \cos \theta _2=B _2 \cos 30^{\circ}=\frac{\sqrt{3}}{2} B _2 \quad …….(ii) $
Also, time period of a magnetic needle is given by
$ \begin{aligned} T & =2 \pi \sqrt{\frac{I}{M B _H}} \quad …….(iii) \\ \therefore \quad T & \propto \sqrt{\frac{1}{B _H}} \text { or } \frac{T _1}{T _2}=\sqrt{\frac{B _{H _2}}{B _{H _1}}} \quad …….(iv) \end{aligned} $
By putting the values from Eqs. (i) and (ii) into Eq. (iv), we get
$\frac{\frac{2}{2}}{2} =\sqrt{\frac{\sqrt{3} \frac{B _2}{2}}{\frac{B _1}{\sqrt{2}}}} \text { or } (\frac{4}{3})^{2}=\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} $
$\Rightarrow \quad \frac{B _1}{B _2} =0.689 \approx 0.7 T$
BETA
