The torque that a uniform magnetic field exerts on a flat (planar) coil is given byT = NBIAsin 0,where N is the number of turns of wire in the coil, B the magnitude of the magnetic field, I the current flowingin the coil, A the area enclosed by the coil, and 0 the angle between the line perpendicular to the plane of thecoil and the direction of the magnetic field. The expression given above was derived for a rectangular coil butit is valid for any planar coil. The area enclosed by the coil can be square, rectangular, circular, ellipsoidal (asin this problem), or any other two-dimensional shape.In the situation shown in the figure redrawn below, the magnetic field is parallel to the plane of the coil.40.0 cmB|-30.0 cm→|dThus, the angle between the magnetic field and the line perpendicular to the plane of the coil is 0 = 90.0°.With a semimajor axis of a =coil is(0.400 m)/2 and a semiminor axis of b = (0.300 m)/2, the area enclosed by theA = nab =Tt m2,and the magnitude of the torque exerted on the coil isT = NBIAsin 0| × 10-4 T)( [m2 )sin 90.0°= 8x 10-4 N ·m.Note that by the right-hand rule, the magnetic force exerted on the left side of the coil is directed out of thepage while the force exerted on the right side is into the page. Thus, an observer looking from c toward dalong the dashed line shown in the diagram sees the coil rotate counterclockwise. Tutorial ExerciseAn eight-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm (seefigure). The coil lies in the plane of the page and carries a clockwise current of 6.17 A. If the coil is in auniform magnetic field of 1.92 x 10-4 T directed toward the left of the page, what is the magnitude of thetorque on the coil? Hint: The area of an ellipse is A = nab, where a and b are, respectively, the semimajor andsemiminor axes of the ellipse.B40.0 cm- 30.0 cm-|

1. Area(A) 2. The magnitude of the torque. given, magnetic fieldB=1.92×10-4 TCurrentI=6.17 Asemi…

The torque that a uniform magnetic field exerts on a flat (planar) coil is given by T = NBIAsin 0, where N is the number of turns of wire in the coil, B the magnitude of the magnetic field, I the current flowir in the coil, A the area enclosed by the coil, and 0 the angle between the line perpendicular to the plane of th coil and the direction of the magnetic field. The expression given above was derived for a rectangular coil bu it is valid for any planar coil. The area enclosed by the coil can be square, rectangular, circular, ellipsoidal (as in this problem), or any other two-dimensional shape. In the situation shown in the figure redrawn below, the magnetic field is parallel to the plane of the coil. 40.0 cm В |-30.0 cm-| d Thus, the angle between the magnetic field and the line perpendicular to the plane of the coil is 0 = 90.0°. With a semimajor axis of a = (0.400 m)/2 and a semiminor axis of b = (0.300 m)/2, the area enclosed by th coil is A = nab = Tt m2, and the magnitude of the torque exerted on the coil is T = NBIAsin 0 ]a m2)sin 90.0° = 8 х 10-4 x 10¬4 N · m. %3D

The torque that a uniform magnetic field exerts on a flat (planar) coil is given by T = NBIAsin 0, where N is the number of turns of wire in the coil, B the magnitude of the magnetic field, I the current flowir in the coil, A the area enclosed by the coil, and 0 the angle between the line perpendicular to the plane of th coil and the direction of the magnetic field. The expression given above was derived for a rectangular coil bu it is valid for any planar coil. The area enclosed by the coil can be square, rectangular, circular, ellipsoidal (as in this problem), or any other two-dimensional shape. In the situation shown in the figure redrawn below, the magnetic field is parallel to the plane of the coil. 40.0 cm В |-30.0 cm-| d Thus, the angle between the magnetic field and the line perpendicular to the plane of the coil is 0 = 90.0°. With a semimajor axis of a = (0.400 m)/2 and a semiminor axis of b = (0.300 m)/2, the area enclosed by th coil is A = nab = Tt m2, and the magnitude of the torque exerted on the coil is T = NBIAsin 0 ]a m2)sin 90.0° = 8 х 10-4 x 10¬4 N · m. %3D

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Magnetic Field Of Coaxial Cable

The word coaxial means same axis. So for a long straight cable to be coaxial, it must have concentric cylindrical shaped conductor around it. Coaxial cable (popularly called ‘coax’) has various applications ranging from current conductors to radiofrequency signal transmitters. This cable has lower emission losses and provides protection from electromagnetic interference which allows signals with less power to transmit over longer distances.For example, currents produced in the pickup of a stereo turntable generally travel to the amplifier through a coaxial cable.The self-inductance of a coaxial cable is another important characteristic, because it influences the propagation of electrical signals in the cable.

Question

100%

The torque that a uniform magnetic field exerts on a flat (planar) coil is given by
T = NBIAsin 0,
where N is the number of turns of wire in the coil, B the magnitude of the magnetic field, I the current flowing
in the coil, A the area enclosed by the coil, and 0 the angle between the line perpendicular to the plane of the
coil and the direction of the magnetic field. The expression given above was derived for a rectangular coil but
it is valid for any planar coil. The area enclosed by the coil can be square, rectangular, circular, ellipsoidal (as
in this problem), or any other two-dimensional shape.
In the situation shown in the figure redrawn below, the magnetic field is parallel to the plane of the coil.
40.0 cm
B
|-30.0 cm→|
d
Thus, the angle between the magnetic field and the line perpendicular to the plane of the coil is 0 = 90.0°.
With a semimajor axis of a =
coil is
(0.400 m)/2 and a semiminor axis of b = (0.300 m)/2, the area enclosed by the
A = nab =
Tt m2,
and the magnitude of the torque exerted on the coil is
T = NBIAsin 0
| × 10-4 T)( [
m2 )sin 90.0°
= 8
x 10-4 N ·m.
Note that by the right-hand rule, the magnetic force exerted on the left side of the coil is directed out of the
page while the force exerted on the right side is into the page. Thus, an observer looking from c toward d
along the dashed line shown in the diagram sees the coil rotate counterclockwise.

Tutorial Exercise
An eight-turn coil encloses an elliptical area having a major axis of 40.0 cm and a minor axis of 30.0 cm (see
figure). The coil lies in the plane of the page and carries a clockwise current of 6.17 A. If the coil is in a
uniform magnetic field of 1.92 x 10-4 T directed toward the left of the page, what is the magnitude of the
torque on the coil? Hint: The area of an ellipse is A = nab, where a and b are, respectively, the semimajor and
semiminor axes of the ellipse.
B
40.0 cm
- 30.0 cm-|

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