Answered: Calculate the maximum emf induced ( in…

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Calculate the maximum emf induced ( in units of mV) in the coil by this field.

University Physics Volume 2

18th Edition

ISBN:9781938168161

Author:OpenStax

Publisher:OpenStax

Chapter13: Electromagnetic Induction

Section: Chapter Questions

Problem 37P: An automobile with a radio antenna 1.0 m long travels at 100.0 km/h in a location where theEarth’s...

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A coil of area 0.100 m2 is rotating at 60.0 rev/s with the axis of rotation perpendicular to a 0.200-T magnetic field. (a) If the coil has 1 000 turns, what is the maximum emf generated in it? (b) What is the orientation of the coil with respect to the magnetic field when the maximum induced voltage occurs?
An automobile with a radio antenna 1.0 m long travels at 100.0 km/h in a location where theEarth’s horizontal magnetic field is 5.5105T . What is the maximum possible emf induced in the antenna due to this motion?
Two long, parallel wires cy equal currents in opposite directions. The radius of each wire is a, and the distance between the centers of the wires is d, Show that if the magnetic flux within the wires themselves can be ignored, the self-inductance of a length 1 of such a pair of wires is L 0 l in d1 a (Hint Calculate the magnetic flux through a rectangle of length 1 between the wires and then use L=N/I
A long solenoid with n= 10 turns per centimeter has a cross-sectional area of 5.0 cm2and carries a current of 0.25 A. A coil with five turns encircles the solenoid. When the current through the solenoid is turned off, it decreases to zero in 0.050 s. What is the average emf induced in the coil?
A thin conducting bar (60.0 cm long) aligned in the positive y direction is moving with velocity v=(1.25m/s)i in a region with a spatially uniform 0.400-T magnetic field directed at an angle of 36.0 above the xy plane. a. What is the magnitude of the emf induced along the length of the moving bar? b. Which end of the bar is positively charged?
The square armature coil of an alternating current generator has 200 turns and is 20.0 cm on side. When it rotates at 3600 rpm, its peak output voltage is 120 V. (a) Wliat is the frequency' of the output voltage? (b) What is the strength of the magnetic field in which the coil is turning?
The accompanying figure shows a metal disk of inner radius r1 and other radius r2 rotating at an angular velocity while in a uniform magnetic field directed parallel to the rotational axis. The brush leads of voltmeter are connected to the dark's inner and outer surfaces as shown. What is the reading of the voltmeter ?
A parallel-plate capacitor with plate separation d is connected to a source of emf that places a time-dependent voltage V(t) across its circular plates of radius r0and area (a) Write an expression for the time rate of change of energy inside the capacitor in terms of V(t) and dV(t)/ dt. (b) Assuming that V(t) is increasing with time, identify the directions of the elecuic field lines inside the capacitor and of the magnetic field lines at the edge of the region between the plates, and then the direction of the Poynting vector S at this location. (c) Obtain expressions for the time dependence of E(t), for B(t) from the displacement current, and for the magnitude of the Poynting vector at the edge of the region between the plates. (d) From S , obtain an expression In terms of ‘(t) and dV(t)/dt for the rate at which electromagnetic field energy the region between the plates. (e) Compare the results of pails (a) and (d) and explain the relationship between them.
The magnetic flux through the loop shown in the accompanying figure varies with time according to m=2.00e3tsin(120t) , where m is in milliwebers. What are the direction and magnitude of the current through the 5.00 resistor at (a) t= 0; (b) t=2.17102 s and (c)t=3.00 s?
A time-dependent uniform magnetic field of magnitude B(t) is confined in a cylindrical region of radius R. A conducting rod of length 2D is placed in the region, as shown below. Show that the emf between the ends of the rod is given by dBdtDR2D2 . ( Hint: To find the between the ends, we need to integrate the electric field from one end to the other. To find the electric field, use Faraday’s law as “Ampere’s law for E”.)
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