Online Homework 7 Solution Video Tutor: Current-Carrying Wire in Magnetic Field Part A The figure shows a wire that is connected to a power supply and suspended between the poles of a magnet. When the switch is closed, the wire deflects in the direction shown. Which of the dashed boxes A–D represents the position of the north magnetic pole? Sol:) 本題電流方向為流出螢幕,再由右手定則即可判斷磁場方向。 So the answer is “C”. 1 Video Tutor: Magnet and Electron Beam Part A The figure shows the path of a charged particle moving in a magnetic field directed into the screen.What is the particle’s charge? Sol:) v v v FB = qv × B q<0 q 小於 0 使得 FB 成為圓周運動之向心力。 So the answer is “negative”. 2 Torque on a Current Loop in a Magnetic Field Part A A current I flows in a plane rectangular current loop with height ω and v horizontal sides b. The loop is placed into a uniform magnetic field B in such v a way that the sides of length ω are perpendicular to B , and there is an angle v θ between the sides of length b and B . Calculate τ , the magnitude of the torque about the vertical axis of the current loop due to the interaction of the current through the loop with the magnetic field. Express the magnitude of the torque in terms of the given variables. You will need a trigonomeric function [e.g., sin (θ ) or cos(θ ) ]. Use B for the magnitude of the magnetic field. Sol:) 磁場在長度為 ω 的導線上所造成的力 F1 如下圖。 F1 = IωB b 由 F1 造成的力矩 τ 1 = 2 IωB cos θ 2 磁場在長度為 b 的導線上所造成的力大小相等方向相反,互相抵消。 因此 τ = IbωB cos(θ ) 3 Part B Give a more general expression for the magnitude of the torque τ . Rewrite the answer found in Part A in terms of the magnitude of the magnetic dipole moment of the current loop m. Define the angle between the vector perpendicular to the plane of the coil and the magnetic field to be ψ, noting that this angle is the complement of angle θ in Part A. Give your answer in terms of the magnetic moment m, magnetic field B, and ψ. Sol:) v v v The more general vector form of this expression is τ = m × B v v 其中, m = IA So τ = mB sin (φ ) Part C A current I flows around a plane circular loop of radius r, giving the loop a magnetic dipole moment of magnitude m. The loop is placed in a v uniform magnetic field B , with an angle ψ between the direction of the field lines and the magnetic dipole moment as shown in the figure. Find an expression for the magnitude of the torque τ on the current loop. Express the torque explicitly in terms of r, I, π, ψ, and B (where τ and B are the magnitudes of the respective vector quantities). Do not use m. You will need a trigonomeric function [e.g., sin (θ ) or cos(θ ) ]. Sol:) Area of circle A = r 2π v v v τ = m× B τ = mB sin (φ ) = Ir 2πB sin (φ ) 4 Problem 27.72 Zeeman effect. In the Bohr model of the hydrogen atom, the electron is held in its circular orbit of radius τ about its proton nucleus by electrostatic attraction. If the atoms are placed in a weak magnetic field v B , the rotation frequency of electrons rotating in a plane perpendicular v to B is changed. v Part A Find ∆f , assuming the force due to B is much less than that due to electrostatic attraction of the nucleus. Express your answer in terms of the variables B, m, e, and appropriate constants. If there is more than one answer, separate them by a comma. Sol:) 未加磁場前,電子繞原子核做圓周運動之向心加速度來自於電子與原子核間的吸 引力。 (r ⋅ 2πf 0 ) ke 2 v2 = m = m r r r2 2 ke ⇒ f0 = 2 3 4π mr 2 v = rω ω = 2 πf 加磁場後 ke 2 (r ⋅ 2πf ) ⇒ f 2 m qB f - f 2 = 0 ± q(r 2πf )B = m 0 2 r 2πm r 2 f = f 0 + ∆f ( f 0 + ∆ f )2 m qB ( f 0 + ∆f ) - f 02 = 0 2πm qB qB ⇒ f 02 + 2 f 0 ∆f + ∆f 2 m f0 m ∆f - f 02 = 0 2πm 2πm neglect neglect qB ⇒ ∆f = ± 4πm 5 ( ∆f << f 0 ) Problem 27.69 A sort of "projectile launcher" is shown in the figure. A large current moves in a closed loop composed of fixed rails, a power supply, and a very light, almost frictionless bar touching the rails. A 0.80 T magnetic field is perpendicular to the plane of the circuit. Part A If the rails are a distance d = 25 cm apart, and the bar has a mass of 1.4 g, what constant current flow is needed to accelerate the bar from rest to 20 m/s in a distance of 1.4 m? Express your answer to two significant figures and include the appropriate units. Sol:) F = IdB = ma 在 1.4 公尺內加速至 20 m/s,a = 142.86 m/s2 由上式可得 I = 1.0 A Part B In what direction must the field point? Sol:) 由右手定則可判斷 磁場方向為 ”downward” 6