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Chapter 15: Problem 65

Three particles each of mass \(m\) and charge \(q\) are attached to the vertices of a triangular frame, made up of three light rigid rods of equal length \(l\). The frame is rotated at constant angular speed \(\omega\) about an axis perpendicular to the plane of the triangle and passing through its centre. The ratio of the magnetic moment of the system and its angular momentum about the axis of rotation is (A) \(\frac{q}{2 m}\) (B) \(\frac{q}{m}\) (C) \(\frac{2 q}{m}\) (D) \(\frac{4 q}{m}\)

Short Answer

Expert verified
The correct answer is (D) \(\frac{4 q}{m}\).

Step by step solution

01

Calculate the magnetic moment of the system

Since there are three particles moving in a circular path, we can find the magnetic moment of each particle and then add them up to find the total magnetic moment. The magnetic moment '碌' of a single particle moving in a circular path of radius 'r' with charge 'q' and angular speed '蠅' can be found using the formula: \[碌 = I \times A\] Where 'I' is the current, which can be calculated with the formula: \[I = \frac{q \times \omega}{2 \pi}\] and 'A' is the area of the circle formed by the path of the particle: \[A = \pi r^2\] Using these formulas, we can find the magnetic moment for one particle and then multiply by three to get the total.
02

Calculate the angular momentum of the system

The angular momentum 'L' for a single particle with mass 'm', moving in a circle with radius 'r' and angular speed '蠅' can be found using the formula: \[L = m \times r^2 \times \omega\] As there are three particles, we can find the total angular momentum by finding the angular momentum for one particle and then multiplying by three.
03

Finding the Ratio of Magnetic Moment and Angular Momentum

Now that we have calculated the total magnetic moment and the total angular momentum of the system, we can find the ratio of the two using the following formula: \[\frac{碌_{total}}{L_{total}} = \frac{3 \times I \times A}{3 \times m \times r^2 \times \omega}\] We need to find the value of 'r' to be half of the side of the equilateral triangle. Use the relation, \[r = \frac{l}{2 \times \sin(30^\circ)}\] Substituting the values of 'I', 'A', and 'r' in the ratio, simplify and find the answer. The simplified ratio will provide the correct answer choice.

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Most popular questions from this chapter

A straight wire of mass \(200 \mathrm{gm}\) and length \(1.5 \mathrm{~m}\) carries a current of \(2 \mathrm{~A}\). It is suspended in mid-air by a uniform horizontal magnetic field \(B\). The magnitude of \(B\) (in tesla) is $$ \left(g=9.8 \mathrm{~m} / \mathrm{s}^{2}\right) $$ (A) 2 (B) \(1.5\) (C) \(0.55\) (D) \(0.66\)

A proton and an alpha-particle enter a uniform magnetic field with the same velocity. The period of rotation of the alpha particle will be (A) four times that of the proton. (B) two times that of the proton. (C) three times that of the proton. (D) same as that of the proton.

Two long wires are hanging freely. They are joined first in parallel and then in series and they are connected with a battery. In both cases, which type of force acts between the two wires? (A) attraction force when in parallel and repulsion force when in series. (B) repulsion force when in parallel and attraction force when in series. (C) repulsion force in both cases. (D) attraction force in both cases.

A triangular loop of side \(l\) carries a current \(I .\) It is placed in a magnetic field \(B\) such that the plane of the loop is in the direction of \(B\). The torque on the loop is (A) Zero (B) \(I B l^{2}\) (C) \(\frac{\sqrt{3}}{2} I B l^{2}\) (D) \(\frac{\sqrt{3}}{4} I B l^{2}\)

The materials suitable for making electromagnets should have [2004] (A) high retentivity and low coercivity. (B) low retentivity and low coercivity. (C) high retentivity and high coercivity. (D) low retentivity and high coercivity.
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