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Chapter 19: Problem 85

The intrinsic magnetic dipole moment of the electron has magnitude $9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ In other words, the electron acts as though it were a tiny current loop with $N I A=9.3 \times 10^{-24} \mathrm{A} \cdot \mathrm{m}^{2} .$ What is the maximum torque on an electron due to its intrinsic dipole moment in a \(1.0-T\) magnetic field?

Short Answer

Expert verified
Answer: The maximum torque on an electron due to its intrinsic dipole moment in a 1.0-T magnetic field is \(9.3 \times 10^{-24} N\cdot m\).

Step by step solution

01

Write the torque formula related to magnetic dipole moment and magnetic field

We know that the torque τ acting on a magnetic dipole of magnetic moment μ in a magnetic field B is given by τ = μ * B * sin θ, where θ is the angle between the magnetic moment and the magnetic field.
02

Find the maximum torque

To find the maximum torque, we need the maximum value of sin θ, which is 1 when θ = 90 degrees. Therefore, the maximum torque is given by τ_max = μ * B.
03

Substitute the given values and calculate the maximum torque

Given, intrinsic magnetic dipole moment of the electron (μ) = \(9.3 \times 10^{-24} A\cdot m^2\) and magnetic field (B) = 1.0 T. Substitute these values into the formula: τ_max = μ * B = \((9.3 \times 10^{-24} A\cdot m^2) * (1.0 T)\). Hence, τ_max = \(9.3 \times 10^{-24} N\cdot m\). Thus, the maximum torque on an electron due to its intrinsic dipole moment in a 1.0-T magnetic field is \(9.3 \times 10^{-24} N\cdot m\).

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Most popular questions from this chapter

At a certain point on Earth's surface in the southern hemisphere, the magnetic field has a magnitude of \(5.0 \times 10^{-5} \mathrm{T}\) and points upward and toward the north at an angle of \(55^{\circ}\) above the horizontal. A cosmic ray muon with the same charge as an electron and a mass of $1.9 \times 10^{-28} \mathrm{kg}$ is moving directly down toward Earth's surface with a speed of \(4.5 \times 10^{7} \mathrm{m} / \mathrm{s} .\) What is the magnitude and direction of the force on the muon?
A square loop of wire of side \(3.0 \mathrm{cm}\) carries \(3.0 \mathrm{A}\) of current. A uniform magnetic field of magnitude \(0.67 \mathrm{T}\) makes an angle of \(37^{\circ}\) with the plane of the loop. (a) What is the magnitude of the torque on the loop? (b) What is the net magnetic force on the loop?
The strength of Earth's magnetic field, as measured on the surface, is approximately \(6.0 \times 10^{-5} \mathrm{T}\) at the poles and $3.0 \times 10^{-5} \mathrm{T}$ at the equator. Suppose an alien from outer space were at the North Pole with a single loop of wire of the same circumference as his space helmet. The diameter of his helmet is \(20.0 \mathrm{cm} .\) The space invader wishes to cancel Earth's magnetic field at his location. (a) What is the current required to produce a magnetic ficld (due to the current alone) at the center of his loop of the same size as that of Earth's field at the North Pole? (b) In what direction does the current circulate in the loop, CW or CCW, as viewed from above, if it is to cancel Earth's field?
A magnet produces a \(0.30-\mathrm{T}\) field between its poles, directed to the east. A dust particle with charge \(q=-8.0 \times 10^{-18} \mathrm{C}\) is moving straight down at \(0.30 \mathrm{cm} / \mathrm{s}\) in this field. What is the magnitude and direction of the magnetic force on the dust particle?
A long straight wire carries a \(4.70-\) A current in the positive \(x\) -direction. At a particular instant, an electron moving at $1.00 \times 10^{7} \mathrm{m} / \mathrm{s}\( in the positive \)y\( -direction is \)0.120 \mathrm{m}$ from the wire. Determine the magnetic force on the electron at this instant. Sce the figure with Problem \(65 .\)
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