{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

10homework08

# 10homework08 - Husain Zeena Homework 8 Due 4:00 am Inst...

This preview shows pages 1–3. Sign up to view the full content.

Husain, Zeena – Homework 8 – Due: Mar 23 2004, 4:00 am – Inst: Sonia Paban 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before making your selection. The due time is Central time. 001 (part 1 of 1) 10 points A current I = 5 . 54 Amps flows through a wire perpendicular to the paper and towards the reader at A and back in the opposite direction at C . Consider the wires below the plane at A and C to be semi-infinite. In the figure, L 1 = 3 m, R = 3 . 7 m, and L 2 = 3 m and there is an external uniform magnetic field B = 8 . 53 T, which is directed into the paper. Being ’external’, this field is NOT due to the current in the wire. Caution: It may be necessary to take into account the contribution from the long straight wire which runs up to and down from the underneath side of the page. L 1 I I L 2 C A R O B What is the magnitude of the magnetic field at the center of the arc O due to the current in the wire? Correct answer: 2 . 62661 × 10 - 7 T. Explanation: The two straight current segments within the plane of the paper do not contribute to the magnetic field at point O , because they are parallel to the radius vector from that point. Therefore ds × ˆ r = 0 on these segments. The contribution from the curved part of the wire is easy to find using the Biot-Savart law B = μ 0 I ds × ˆ r 4 π r 2 = μ 0 I π 2 4 π R = μ 0 I 8 R ( - ˆ k ) , but one must not forget to take into account the contribution from the long straight wire which runs into and out of the page. For the wire above this is B 1 = μ 0 I 4 π ( L 1 + R ) ˆ ı , while for the wire below we have B 2 = μ 0 I 4 π ( L 2 + R ) ˆ . These three must be added as vectors, giving the magnitude of the field | B O | = μ 0 I 8 R 2 + μ 0 I 4 π ( R + L 1 ) 2 + μ 0 I 4 π ( R + L 2 ) 2 1 / 2 = (8 . 26866 × 10 - 8 T) 2 +(8 . 26866 × 10 - 8 T) 2 +(2 . 35195 × 10 - 7 T) 2 1 2 = 2 . 62661 × 10 - 7 T . 002 (part 1 of 2) 10 points The loop in the figure carries a current 1 . 13 A. The semicircular portion has a radius 5 . 73 cm. A 5 . 73 cm 5 . 73 cm 2 × 5 . 73 cm 1 . 13 A Determine the magnitude of the magnetic field at A. Correct answer: 1 . 17733 × 10 - 5 T. Explanation: Let : I = 1 . 13 A , R = 5 . 73 cm

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Husain, Zeena – Homework 8 – Due: Mar 23 2004, 4:00 am – Inst: Sonia Paban 2 A R R 2R I Basic Concepts: The Biot-Savart Law is given by d B = μ 0 4 π I d × ˆ r r 2 . (1) Solution: Note: The distance from a current element on a circle to the center is a constant, namely r , so we can pull this out of the inte- gral. Also, the current element, I dl , is always perpendicular to ˆ r , so sin θ = 1. Hence, B full circle = μ 0 4 π I r 2 d = μ 0 4 π I r 2 2 π r = μ 0 I 2 r , where a half-circular arc is one-half this value. Assume: Out of the page to be positive. The angle θ (between the vector and the vector r ) is from the Biot-Savart Law, Eq. (1), for the magnetic field B a distance a away from a wire segment B = μ 0 I 4 π a θ 2 θ 1 sin θ d θ = μ 0 I 4 π a (cos θ 1 - cos θ 2 ) , where a = R for the sides of the semi-square in figure.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}