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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?
The conversion between atomic mass units and kilograms is $$1 \mathrm{u}=1.66 \times 10^{-27} \mathrm{kg}$$ Natural carbon consists of two different isotopes (excluding $^{14} \mathrm{C},$ which is present in only trace amounts). The isotopes have different masses, which is due to different numbers of neutrons in the nucleus; however, the number of protons is the same, and subsequently the chemical properties are the same. The most abundant isotope has an atomic mass of \(12.00 \mathrm{u} .\) When natural carbon is placed in a mass spectrometer, two lines are formed on the photographic plate. The lines show that the more abundant isotope moved in a circle of radius \(15.0 \mathrm{cm},\) while the rarer isotope moved in a circle of radius \(15.6 \mathrm{cm} .\) What is the atomic mass of the rarer isotope? (The ions have the same charge and are accelerated through the same potential difference before entering the magnetic field.)
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(a) A proton moves with uniform circular motion in a magnetic field of magnitude 0.80 T. At what frequency \(f\) does it circulate? (b) Repeat for an electron.
A proton moves in a helical path at speed \(v=4.0\) $\times 10^{7} \mathrm{m} / \mathrm{s}$ high above the atmosphere, where Earth's magnetic field has magnitude \(B=1.0 \times 10^{-6} \mathrm{T}\). The proton's velocity makes an angle of \(25^{\circ}\) with the magnetic field. (a) Find the radius of the helix. [Hint: Use the perpendicular component of the velocity.] (b) Find the pitch of the helix-the distance between adjacent "coils." [Hint: Find the time for one revolution; then find how far the proton moves along a field line during that time interval.]
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