# HW8c - GenericStudent – Homework 8 – savrasov – 39823...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: GenericStudent – Homework 8 – savrasov – 39823 – Jan 02, 2008 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 2) 1 points Consider the application of the Biot-Savart law Δ ~ B = μ 4 π I Δ ~ ‘ × ˆ r r 2 . Consider the magnetic field at O due to the current segments B B +semicircle+ CC . The length of the linear segments, B B = CC = d . The semicircle has a radius 3 m. The permeability of free space is 1 . 25664 × 10- 6 T · m / A . x y 7 A 7 A 7 A 180 ◦ O 3 m B B C C A What is the path of the magnetic field B O at the origin O due to the 7 A current? 1. into the page 2. out of the page correct 3. undetermined, since B = 0 Explanation: Basic Concepts: Biot-Savart Law Solution: ˆ r is pointing from A to O ; i.e. , from the location of the source element to the location of the field of concern. So- ˆ is the direction. 002 (part 2 of 2) 1 points Find the magnitude of the magnetic field due to the current segments described above at O . Correct answer: 7 . 33038 × 10- 7 T. Explanation: Let : μ = 4 π × 10- 7 T · m / A = 1 . 25664 × 10- 6 T · m / A , I = 7 A , and r = 3 m . Since the Biot-Savart equation has the cross product ˆ ı × ˆ r in the numerator, this term implies that the contribution of B B and CC to the field at O is zero. The magnetic field at at the center of an arc with a current I is B = μ I 4 π Z d~s × ˆ r r 2 = μ I 4 π r 2 Z ds = μ I 4 π r 2 Z r dθ = μ I 4 π r Z π dθ = μ I 4 π r θ fl fl fl fl π = μ I 4 r = (1 . 25664 × 10- 6 T · m / A) (7 A) 4 (3 m) = 7 . 33038 × 10- 7 T . keywords: 003 (part 1 of 3) 1 points A conductor in the shape of a square, whose sides are of length 0 . 802 m, carries a clockwise 14 . 7 A current as shown in the figure below. The permeability of free space is 1 . 25664 × 10- 6 T · m / A. . 802 m 14 . 7 A P GenericStudent – Homework 8 – savrasov – 39823 – Jan 02, 2008 2 The permeability of free space is 1 . 25664 × 10- 6 T · m / A. What is the magnitude of the magnetic field at point P (at the center of the square loop) due to the current in the wire? Correct answer: 20 . 7371 μ T. Explanation: Let : μ = 1 . 25664 × 10- 6 T · m / A , ‘ = 0 . 802 m , and I = 14 . 7 A . By the Biot-Savart law, dB = μ 4 π I d~s × ˆ r r 2 . Consider a thin, straight wire carring a con- stant current I along the x-axis with the y- axis pointing towards the center of the square, as in the following figure. y x O r P x I ˆ r ds a θ Let us calculate the total magnetic field at the point P located at a distance a from the wire. An element d~s is at a distance r from P . The direction of the field at P due to this element is out of the paper, since d~s × ˆ r is out of the paper. In fact, all elements give a contribution directly out of the paper at P . Therefore, we have only to determine the magnitude of the field at...
View Full Document

## This note was uploaded on 01/10/2010 for the course PHY na taught by Professor Richardscalettar during the Winter '09 term at UC Davis.

### Page1 / 7

HW8c - GenericStudent – Homework 8 – savrasov – 39823...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online