Problem Set 7 Solutions

Problem Set 7 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2007 Problem Set 7 Solutions Problem 1 Quickies a Two

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

View Full Document Right Arrow Icon
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2007 Problem Set 7 Solutions Problem 1: Quickies… a) Two semicircular arcs have radii R 2 and R 1 , carry current i , and share the same center of curvature C. What is the magnitude of the net magnetic field at C? The inner semi-circle makes a field into the page, the outer one out of the page. The inner is closer and hence stronger, and hence the net field is into the page. In class you calculated the magnetic field from a semi-circle of radius R to be 0 4 B iR μ = . So: 0 11 into the page 4 i RR ⎛⎞ =− ⎜⎟ B G 12 ⎝⎠ le, what is θ ? b) A wire with current i is shown at left. Two semi-infinite straight sections, both tangent to the same circle with radius R , are connected by a circular arc that has a central angle and runs along the circumference of the circle. The connecting arc and the two straight sections all lie in the same plane. If B = 0 at the center of the circ The straight portions both make a field out of the page at the center of the circle while the arc makes one into the page. These must be equal so that the fields cancel. Two semi- infinite lines together make an infinite line, and we calculated (using Ampere’s law) that the field from an infinite wire is 0 2 B = π . The arc is just a fraction of a circle so it creates a fraction of the field that a whole circle does at its center: () 0 22 Bi R = θπ . For these to be equal we must have 00 24 2 r a d i a n s RiR μμ = θ π θ = c) The figure at left shows two closed paths wrapped around two conducting loops carrying currents i 1 and i 2 . What is the value of the integral for (a) path 1 and (b) path 2? To do this you have to use the right hand rule to check whether the currents are positive or negative relative to the path. On path 1 i 1 penetrates in the negative direction while i 2 penetrates in the positive direction, so 21 o di i Bs G G v . On path 2 i 1 penetrates twice in the negative direction and i 2 once in the negative direction so 2 o i ⋅= μ + G G v PS07 Solutions-1
Background image of page 1

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

View Full DocumentRight Arrow Icon
Problem 2: Two Currents In the figure at left a long circular pipe with outside radius R carries a (uniformly distributed) current i into the page. A wire runs parallel to the pipe at a distance of 3.00 R from center to center. Find the current in the wire such that the ratio of the magnitude of the net magnetic field at point P to the magnitude of the net magnetic field at the center of the pipe is x , but it has the opposite direction. The field at point P is due both to the pipe and the wire. The field at the center of the pipe is ONLY due to the wire. Since the direction of these two is opposite the current in the wire must create an opposite direction field from the pipe at point P and hence it must also be into the page.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/07/2008 for the course 8 8.02 taught by Professor Hudson during the Fall '07 term at MIT.

Page1 / 10

Problem Set 7 - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2007 Problem Set 7 Solutions Problem 1 Quickies a Two

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online