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Chapter 15: Problem 3

Three identical bar magnets, each of magnetic moment \(M\), are placed in the form of an equilateral triangle with north pole of one touching the south pole of the other as shown. The net magnetic moment of the system is (A) Zero (B) \(3 M\) (C) \(\frac{3 M}{2}\) (D) \(M \sqrt{3}\)

Short Answer

Expert verified
The net magnetic moment of the system is zero because the x and y components of the individual magnetic moments sum up to zero. Therefore, the correct answer is (A) Zero.

Step by step solution

01

Understand the magnetic moment properties

Magnetic moment \(M\) is a vector quantity, with its direction pointing from the south pole to the north pole of the magnet. In the equilateral triangle formation, the three magnetic moments will have different directions which are \(120^{\circ}\) apart.
02

Set up the coordinate system and assign vectors

Let's set up a coordinate system where the origin is at the center of the equilateral triangle and the x-axis is parallel to the base of the triangle. We have three magnetic moments: - \(M_1\) will be assigned at 0 degrees to the x-axis. - \(M_2\) will be 120 degrees to the x-axis. - \(M_3\) will be 240 degrees to the x-axis. Now, we can break down these magnetic moments into their x and y components.\( M_{1x} = M \cos{(0)} = M \) and \( M_{1y} = M \sin{(0)} = 0 \) \( M_{2x} = M \cos{(120)} = -\frac{1}{2} M \) and \( M_{2y} = M \sin{(120)} = \frac{\sqrt{3}}{2} M \) \( M_{3x} = M \cos{(240)} = -\frac{1}{2} M \) and \( M_{3y} = M \sin{(240)} = -\frac{\sqrt{3}}{2} M \)
03

Compute the net magnetic moment

To find the net magnetic moment, we can sum up the x and y components of the individual magnetic moments. \( M_{net_x} = M_{1x} + M_{2x} + M_{3x} = M - \frac{1}{2} M - \frac{1}{2} M = 0 \) \( M_{net_y} = M_{1y} + M_{2y} + M_{3y} = 0 + \frac{\sqrt{3}}{2} M - \frac{\sqrt{3}}{2} M = 0\) Since both the x and y components of the net magnetic moment are zero, the net magnetic moment is also zero. Thus, the correct answer is (A) Zero.

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Most popular questions from this chapter

The region between \(X=0\) and \(X=L m\) is filled with uniform steady magnetic field \(2 \mathrm{~T} \hat{k}\). A particle of mass \(2 \mathrm{~kg}\), positive charge \(1 \mathrm{C}\) and velocity \(2(\mathrm{~m} / \mathrm{s}) \hat{i}\) travels along \(x\)-axis and enters the region of the magnetic field (neglect gravity). Find the value of \(L\) if the particle emerges from the region of magnetic field with its final velocity at an angle \(30^{\circ}\) to its initial velocity.

Ratio of radii of paths when an electron and a proton enters at right angles to a uniform field with (A) same velocity is \(\frac{1}{1840}\). (B) same momentum is 1 . (C) same kinetic energy is \(\frac{1}{43}\). (D) same kinetic energy is 43 .

A particle of charge \(-16 \times 10^{-18} \mathrm{C}\) moving with velocity \(10 \mathrm{~m} / \mathrm{s}\) along the \(x\)-axis enters a region where a magnetic field of induction \(B\) is along the \(y\)-axis and an electric field of magnitude \(10^{4} \mathrm{~V} / \mathrm{m}\) is along the negative \(z\)-axis. If the charged particle continues moving along the \(x\)-axis, the magnitude of \(B\) is (A) \(10^{3} \mathrm{~Wb} / \mathrm{m}^{2}\) (B) \(10^{5} \mathrm{~Wb} / \mathrm{m}^{2}\) (C) \(10^{16} \mathrm{~Wb} / \mathrm{m}^{2}\) (D) \(10^{-3} \mathrm{~Wb} / \mathrm{m}^{2}\)

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) \(\vec{F}_{1}=\vec{F}_{2}=0\).

If in a circular coil \(A\) of radius \(R\), current \(I\) is flowing and in another coil \(B\) of radius \(2 R\) a current \(2 I\) is flowing, then the ratio of the magnetic fields \(B_{A}\) and \(B_{B}\) produced by them will be (A) 1 (B) 2 (C) \(\frac{1}{2}\) (D) 4
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