/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 If a magnet is suspended at an a... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 72

If a magnet is suspended at an angle of \(30^{\circ}\) to the magnetic meridian, the dip needle makes an angle of \(45^{\circ}\) with the horizontal. The real dip is (A) \(\tan ^{-1}\left(\frac{\sqrt{3}}{2}\right)\) (B) \(\tan ^{-1}(\sqrt{3})\) (C) \(\tan ^{-1}\left(\sqrt{\frac{3}{2}}\right)\) (D) \(\tan ^{-1} \frac{2}{\sqrt{3}}\)

Short Answer

Expert verified
The real dip angle is (D) \(\tan ^{-1} \frac{2}{\sqrt{3}}\).

Step by step solution

01

Understand the problem's given values

In this problem, we are given that the magnet is suspended at an angle of 30掳 to the magnetic meridian, and the dip needle makes an angle of 45掳 with the horizontal.
02

Use the tangent formula for real dip

The real dip angle \(\alpha\) can be found using the tangent formula: \[\tan(\alpha)=\frac{\tan(45^\circ)}{\cos(30^\circ)}\]
03

Calculate the values for the formula

Now, let's find the values for that formula: \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\) \(\tan(45^\circ) = 1\)
04

Plug the values into the formula

Now we have the necessary values to plug into the formula. \[\tan(\alpha) = \frac{1}{\frac{\sqrt{3}}{2}}\]
05

Simplify and find the real dip angle

Simplify the equation to find the value of \(\tan(\alpha)\): \[\tan(\alpha) = \frac{2}{\sqrt{3}}\] Now, find the inverse tangent of this value to get the real dip angle: \(\alpha = \tan^{-1}\left(\frac{2}{\sqrt{3}}\right)\) From the given options, the real dip angle is: (D) \(\tan ^{-1} \frac{2}{\sqrt{3}}\)

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Most popular questions from this chapter

A long solenoid has 200 turns per \(\mathrm{cm}\) and carries a current \(i\). The magnetic field at its centre is \(6.28 \times 10^{-2}\) \(\mathrm{Wb} / \mathrm{m}^{2}\) Another long solenoid has 100 turns per \(\mathrm{cm}\) and it carries a current \(\frac{i}{3} .\) The value of the magnetic field at its centre is (A) \(1.05 \times 10^{-2} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(1.05 \times 10^{-5} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(1.05 \times 10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(1.05 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^{2}\)

A bar magnet has a magnetic moment of \(2.5 \mathrm{JT}^{-1}\) and is placed in a magnetic field of \(0.2 \mathrm{~T}\). Work done in turning the magnet from parallel to anti-parallel position relative to field direction is (A) \(0.5 \mathrm{~J}\) (B) \(1 \mathrm{~J}\) (C) \(2 \mathrm{~J}\) (D) \(0 \mathrm{~J}\)

Two identical conducting wires \(A O B\) and \(C O D\) are placed at right angles to each other. The wire \(A O B\) carries an electric current \(I_{1}\) and \(C O D\) carries a current \(I_{2} .\) The magnetic field on a point lying at a distance \(d\) from \(O\), in a direction perpendicular to the plane of the wires \(A O B\) and \(C O D\), will be given by [2007] (A) \(\frac{\mu_{0}}{2 \pi d}\left(I_{1}^{2}+I_{2}^{2}\right)\) (B) \(\frac{\mu_{0}}{2 \mu}\left(\frac{I_{1}+I_{2}}{d}\right)^{\frac{1}{2}}\) (C) \(\frac{\mu_{0}}{2 \pi d}\left(I_{1}^{2}+I_{2}^{2}\right)^{\frac{1}{2}}\) (D) \(\frac{\mu_{0}}{2 \pi d}\left(I_{1}+I_{2}\right)\)

A metal disc of radius \(R=6 \mathrm{~cm}\) is mounted on a frictionless axle. The current can flow through the axle out along the disc to a sliding contact of rim of the disc. A uniform magnetic field \(B=2 \mathrm{~T}\) is parallel to the axis of the disc. When the current is \(3 \mathrm{~A}\), the disc rotates with constant angular velocity. The frictional force at the rim between the stationary electrical contact and the rotating rim is \(9 x \times 10^{-2} \mathrm{~N}\). Find the value of \(x\).

Two coaxial solenoids of different radii carry current \(I\) in the same direction. Let \(\vec{F}_{1}\) be the magnetic force on the inner solenoid due to the outer one and \(\vec{F}_{2}\) be the magnetic force on the outer solenoid due to the inner one. Then (A) \(\vec{F}_{1}\) is radially inwards and \(\vec{F}_{2}\) is radially outwards. (B) \(\vec{F}_{1}\) is radially inwards and \(\vec{F}_{2}=0\). (C) \(\vec{F}_{1}\) is radially outwards and \(\vec{F}_{2}=0\). (D) \(\vec{F}_{1}=\vec{F}_{2}=0\).
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