Magnetic and Electric Fields: Induced EMF and Current Flow - Prof. Leo E. Piilonen, Exams of Physics

Download Magnetic and Electric Fields: Induced EMF and Current Flow - Prof. Leo E. Piilonen and more Exams Physics in PDF only on Docsity! PHYS 2306 Test #3 Solutions 1. Electrons flow along a wire segment that is near the North pole of a permanent magnet. The magnetic force pushes the wire segment Electron flow √ 1© upward. 2© downward. 3© in the direction of the electron flow. 4© away from the magnet. 5© toward the magnet. The current I flows in the direction opposite the depicted arrow since the flowing charges are neg- ative. The magnetic field points out of the magnet’s North pole. The force vector is ~F = ~Iℓ× ~B on some length ℓ of the wire. 2. The gap between the coaxial circular plates of a gigantic parallel-plate capacitor contains a uni- form electric field that is changing with time as the plates are charged. The plates have radius R and are separated by distance ℓ. If an observer is at radius r < R and midway between the plates, she observes an induced magnetic field that is 1© proportional to 1/r. 2© proportional to 1/r2. 3© proportional to r2.√ 4© proportional to r. 5© zero. According to Ampère’s Law as extended by Maxwell, the induced magnetic field in the gap is related to the rate of change of electric field via ∮ ~B · d~ℓ = (µ0ε0)(dΦE/dt) or (B)(2πr) = (µ0ε0)(πr 2)(dE/dt). Solving for the induced magnetic field gives B = 1 2 (πµ0ε0r)(dE/dt), which is proportional to r. 3. An infinitely long hollow solenoid of radius R contains a uniform magnetic field B within its interior, parallel to its axis. If the magnitude of this field is changing at a constant rate Ḃ, what is the magnitude of the electric field induced at radius 2R from the solenoid’s axis? √ 1© 1 4 RḂ 2© 1 2 RḂ 3© RḂ 4© 2RḂ 5© zero According to Faraday’s Law, the induced electric field is related to the change of magnetic field via ∮ ~E ·d~ℓ = −dΦB/dt or (E)(2πr) = −(πR2)(Ḃ), where r = 2R for the observer. (The magnetic field is contained within the radius R, i.e., inside the solenoid; this is why I use two different radii here.) Solving for the electric field gives E = 1 4 RḂ. B v 4. A rectangular loop of (resistive) wire is pushed at constant velocity ~v from a field-free region into a region with uniform magnetic field B and then into another field-free region, as shown. Which graph correctly shows the induced current i in the loop as a function of time t? ( = +) Answer: √ 4© t i t i t i 1 2 3 4 5 i t i t As soon as the loop starts entering the region of magnetic field, the flux of magnetic field through it increases linearly with time (because the overlap area increases linearly with time). Therefore, the induced emf in the loop is a constant value, equal to the rate of increase of magnetic flux: Eind = −dΦB/dt. After the loop is entirely within the field region, the flux is constant so its rate of change is zero and the induced emf is zero. Then, while the loop is leaving the field region, the magnetic flux decreases linearly with time, and the emf equals its rate of decrease. (Curve 5© shows the flux ΦB as a function of time; curve 4© is the slope of this flux. You need opposite polarity for the emf on entrance and exit, so you can eliminate curves 1© and 2© as possibilities.) 5. One of Maxwell’s equations contains the term ∮ ~B · d~ℓ. The “o” symbol in the integral sign 1© tells you to integrate over a closed surface. 2© is the same as the subscript in the magnetic permeability µ0. 3© tells you to integrate counterclockwise around the path. 4© tells you to integrate clockwise around the path.√ 5© tells you to integrate around a closed path. What more can I say? 6. Protons of mass m and charge e are first accelerated from rest through a potential difference ∆V and then deflected by a uniform magnetic field B that is perpendicular to their velocity. The radius of the resulting proton trajectory is 1© Bm √ 2 ∆V /e. 2© B √ 2 m∆V/e.√ 3© √ 2 m∆V/e/B. 4© B √ 2 e∆V /m. 5© √ 2 e∆V/m/B. Just after the acceleration, the electrostatic potential energy e∆V has been transformed into the proton’s kinetic energy 1 2 mv2, so the speed is v = √ 2e∆V/m. In the magnetic field region, the proton’s cyclotron radius is R = (mv)/(eB) = (m √ 2e∆V/m)/(eB) = √ 2 m∆V/e/B. (You could do dimensional analysis to find the one answer with the correct units, but that is not so easy when you have these oddball SI units.) field is ~I × r̂, where r̂ points from the segment to P . This direction is easiest to determine using a right-hand rule: point your thumb along the current and your fingers will curl in the direction of the magnetic field—into the page for points like P that are below the segment. 14. A radio receiver is tuned by adjusting the capacitor of an LC circuit. If the capacitance is C0 for a frequency of 600 kHz, then to retune to a frequency of 1200 kHz requires the adjustment of the capacitance to 1© √ 2 C0. 2© 4C0. 3© 2C0.√ 4© C0/4. 5© C0/2. The angular frequency of the LC circuit is ω = 1/ √ LC (which means that the regular fre- quency is f = 1/(2π √ LC, but you don’t need this explicitly). If the frequency is to double, then the capacitance is reduced by a factor of four. (It appears in the denominator’s square root.) 15. 30 15 10 12 6 All resistances in ohms 7 The potential difference across the 6 Ω resistor in this figure is 48 V. What is the current I that flows through the 7 Ω resistor at the bottom? √ 1© 12 A 2© 8 A 3© 6 A 4© 4 A 5© 3 A Applying KVL around the upper right loop, the potential across the 12 Ω resistor must also be 48 V. By KCL, the current through the 7 Ω resistor is then the sum of the currents through the 6 Ω and 12 Ω resistors: I7 = (48/6) + (48/12) = 12 A. 16. An electron and a proton each travel at the same speed around circular orbits in the plane of this page because of the influence of a uniform external magnetic field pointing into the page. Then the electron travels around the circle and the proton travels around the circle (as you view this page from above). 1© , larger, , smaller 2© , larger, , smaller 3© , larger, , smaller 4© , smaller, , larger√ 5© , smaller, , larger The cyclotron radius is R = (mv)/(qB). The proton mass is higher than the electron’s, so the proton’s orbit has the larger radius. If a proton is moving rightward at one instant, the magnetic force ~F = e~v × ~B points upward in the plane of the page by the right-hand rule for the cross product. That is, the proton at “six o’clock” will be steered toward “five o’clock” (counterclockwise). The negative electron travels clockwise. 17. By itself, a magnetic field cannot 1© change the direction of motion of a charged particle.√ 2© change the kinetic energy of a charged particle. 3© change the momentum of a charged particle. 4© change the velocity of a charged particle. 5© exert a force on a charged particle. The magnetic force ~F = q~v × ~B will change the direction of the vectors ~v (velocity) and ~p (momentum) and will exert a force on a moving charged particle. This leaves only 2© as the answer. (In other words, the magnetic force doesn’t change the speed of the moving charged particle, so the kinetic energy—a scalar—is unchanged by this force.) 18. The diagrams show three planar circuits consisting of concentric circular arcs of radii r or 2r or 3r connected by straight radial segments. Each circuit carries current i. Rank the circuits according to the magnitude of the net magnetic field observed at point P , LEAST to GREATEST. 2rP P P A B r 3r C 1© B, then A, then C 2© B, then C, then A 3© A, then C, then B√ 4© C, then B, then A 5© A, then B, then C The magnetic field due to each arc of current is inversely proportional to the radius of the arc. The direction of the field at P from the smaller arc is in same as direction of the field at P from the larger arc in case (A), but opposite in cases (B) and (C), so case (A) must have the strongest net field. The partial cancellation in better in case (C) because one quarter of the small arc is closer to the radius of the large arc, so (C) has the smallest net field at P . 19. Two parallel wires carrying equal currents attract each other with a force F . If both currents are doubled and the wires are moved twice as far apart, the force of attraction will be 1© F/2. 2© F/4. 3© F . 4© 4F .√ 5© 2F . The magnetic force between a common length ℓ of the two wires is F = (µ0I 2ℓ)/(2πr). Double I (which is squared in the numerator) and double r (in the denominator), leaving everything else unchanged. 20. A copper hoop stands in a vertical east-west plane in a region of space where there is a uniform external magnetic field pointing horizontally from south to north. The largest induced emf in the hoop is produced when the hoop is S W E N Up Down 1© moved upward rapidly without spinning. 2© moved westward rapidly without spinning. 3© moved northward rapidly without spinning.√ 4© spun about an east–west axis. 5© spun about a north–south axis. The magnetic flux through the hoop does not change for four of the five choices, so the in- duced emf is zero for them. Only in choice 4© does the hoop experience an induced emf, since the magnetic flux ΦB = ~B · ~A = (B)(πr2) cos θ changes with the rotation angle θ.