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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 has a 5.82-A current flowing clockwise around it. If the coil is in a uniform magnetic field of 1.99 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 = ab, where a and b are, respectively, the semimajor and semiminor axes of the ellipse. O 30.0 cm 40.0 cm Step 1 The magnitude of the torque that a uniform magnetic field exerts on a flat (planar) coil is given by T = μ x B = NIAB sin 0, B where N is the number of turns of wire in the coil, I the current in the coil, A the area enclosed by the coil, B the magnitude of the magnetic field, 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 in the textbook 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. -30.0 cm- Ⓡ 40.0 cm Therefore, the angle between the magnetic field and the magnetic moment of the coil is = 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 the coil is A = ab = 0.400 X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m², and the magnitude of the torque exerted on the coil is T = NIAB sin 0 xm²)( A)( x 10-4 N. m. Submit Skip (you cannot come back) x 10-4 4 T) s sin 90.0⁰ 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. Therefore, 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 has a 5.82-A current flowing clockwise around it. If the coil is in a uniform magnetic field of 1.99 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 = ab, where a and b are, respectively, the semimajor and semiminor axes of the ellipse. O 30.0 cm 40.0 cm Step 1 The magnitude of the torque that a uniform magnetic field exerts on a flat (planar) coil is given by T = μ x B = NIAB sin 0, B where N is the number of turns of wire in the coil, I the current in the coil, A the area enclosed by the coil, B the magnitude of the magnetic field, 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 in the textbook 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. -30.0 cm- Ⓡ 40.0 cm Therefore, the angle between the magnetic field and the magnetic moment of the coil is = 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 the coil is A = ab = 0.400 X Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m², and the magnitude of the torque exerted on the coil is T = NIAB sin 0 xm²)( A)( x 10-4 N. m. Submit Skip (you cannot come back) x 10-4 4 T) s sin 90.0⁰ 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. Therefore, an observer looking from c toward d along the dashed line shown in the diagram sees the coil rotate counterclockwise.

College Physics
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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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 has a 5.82-A current flowing clockwise around it. If the coil is in a uniform magnetic field of 1.99 × 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 = лab, where a and b are, respectively, the semimajor and semiminor axes of the ellipse. 100 +30.0 cm Step 1 The magnitude of the torque that a uniform magnetic field exerts on a flat (planar) coil is given by T = μ x B = NIAB sin 0, where N is the number of turns of wire in the coil, I the current in the coil, A the area enclosed by the coil, B the magnitude of the magnetic field, 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 in the textbook 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. B = 8 40.0 cm 01 +30.0 cm- 40.0 cm Therefore, the angle between the magnetic field and the magnetic moment of the coil is = 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 the coil is A = πab = 0.400 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m², and the magnitude of the torque exerted on the coil is T = NIAB sin 0 ™m²)( A)(! x 10-4 N. m. x 10 Submit Skip (you cannot come back) +T) sin 90.0° 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. Therefore, an observer looking from c toward d along the dashed line shown in the diagram sees the coil rotate counterclockwise.
Transcribed Image Text: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 has a 5.82-A current flowing clockwise around it. If the coil is in a uniform magnetic field of 1.99 × 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 = лab, where a and b are, respectively, the semimajor and semiminor axes of the ellipse. 100 +30.0 cm Step 1 The magnitude of the torque that a uniform magnetic field exerts on a flat (planar) coil is given by T = μ x B = NIAB sin 0, where N is the number of turns of wire in the coil, I the current in the coil, A the area enclosed by the coil, B the magnitude of the magnetic field, 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 in the textbook 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. B = 8 40.0 cm 01 +30.0 cm- 40.0 cm Therefore, the angle between the magnetic field and the magnetic moment of the coil is = 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 the coil is A = πab = 0.400 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. m², and the magnitude of the torque exerted on the coil is T = NIAB sin 0 ™m²)( A)(! x 10-4 N. m. x 10 Submit Skip (you cannot come back) +T) sin 90.0° 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. Therefore, an observer looking from c toward d along the dashed line shown in the diagram sees the coil rotate counterclockwise.
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