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Chapter 13: Problem 1976

A dip needle vibrates in the vertical plane perpendicular to the magnetic meridian. The time period of vibration is found to be \(2 \mathrm{sec}\). The same needle is then allowed to vibrate in the horizontal plane and the time period is again found to be \(2 \mathrm{sec}\). Then the angle of dip is (a) \(0^{\circ}\) (b) \(30^{\circ}\) (c) \(45^{\circ}\) (d) \(90^{\circ}\)

Short Answer

Expert verified
The angle of dip is (c) \(45^{\circ}\).

Step by step solution

01

Formula for time period in vertical plane

In the vertical plane (perpendicular to magnetic meridian), the time period formula of vibration is given by the following relation: \(T_v = 2\pi\sqrt{\frac{I}{mH_v}}\) Where \(T_v\) is the time period in the vertical plane, \(I\) is the moment of inertia of the dip needle, \(m\) is the needle's magnetic moment, and \(H_v\) is the vertical component of the earth's magnetic field.
02

Formula for time period in horizontal plane

In the horizontal plane, the time period formula of vibration is given by the following relation: \(T_h = 2\pi\sqrt{\frac{I}{mH_h}}\) Where \(T_h\) is the time period in the horizontal plane and \(H_h\) is the horizontal component of the earth's magnetic field.
03

Given values of time periods

We are given that the time periods in both planes are the same, i.e., \(T_v = T_h = 2\) seconds. This means that: \(2\pi\sqrt{\frac{I}{mH_v}} = 2\pi\sqrt{\frac{I}{mH_h}}\)
04

Simplifying the equation

By simplifying the equation, we get: \(\frac{1}{\sqrt{H_v}} = \frac{1}{\sqrt{H_h}}\) Squaring both sides, we obtain: \(\frac{1}{H_v} = \frac{1}{H_h}\)
05

Using the angle of dip in magnetic field components

The angle of dip (\(\delta\)) is related to the magnetic field components as follows: \(H_v = H\sin\delta\) \(H_h = H\cos\delta\) Where H is the total magnetic field at that point.
06

Substituting values into the equation

By substituting the expressions for \(H_v\) and \(H_h\) in terms of the angle of dip, we get: \(\frac{1}{H\sin\delta} = \frac{1}{H\cos\delta}\)
07

Simplifying and solving for the angle of dip

By simplifying and solving for the angle of dip (\(\delta\)), we obtain: \(\sin\delta = \cos\delta\) Since \(\sin\delta = \cos\delta\), we can rewrite the equation as \(\sin\delta = \sin\left(\frac{\pi}{2} - \delta\right)\). By using the property \(\sin\alpha=\sin\beta\), we get: \(\delta = \frac{\pi}{2} - \delta\), this results in \(2\delta = \frac{\pi}{2}\), which implies: \(\delta = \frac{\pi}{4}\) Thus, the angle of dip (\(\delta\)) is \(\frac{\pi}{4}\) radians, or \(45^{\circ}\). Therefore, the correct answer is: (c) \(45^{\circ}\)

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Most popular questions from this chapter

A long horizontal wire " \(\mathrm{A}^{\prime \prime}\) carries a current of $50 \mathrm{Amp}$. It is rigidly fixed. Another small wire "B" is placed just above and parallel to " \(\mathrm{A}^{\prime \prime}\). The weight of wire- \(\mathrm{B}\) per unit length is \(75 \times 10^{-3}\) Newton/meter and carries a current of 25 Amp. Find the position of wire \(B\) from \(A\) so that wire \(B\) remains suspended due to magnetic repulsion. Also indicate the direction of current in \(B\) w.r.t. to \(A\).(a) \((1 / 2) \times 10^{-2} \mathrm{~m}\); in same direction (b) \((1 / 3) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction (c) \((1 / 4) \times 10^{-2} \mathrm{~m}\); in same direction (d) \((1 / 5) \times 10^{-2} \mathrm{~m}\); in mutually opposite direction

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