863_Physics ProblemsTechnical Physics

# 863_Physics ProblemsTechnical Physics - Chapter 30*P30.52...

This preview shows page 1. Sign up to view the full content.

Chapter 30 203 *P30.52 At a point at distance x from the left end of the bar, current I 2 creates magnetic field B = + µ π 02 22 2 I hx to the left and above the horizontal at angle θ where tan = x h . This field exerts force on an element of the rod of length dx dI I Id x IIdx x FB = + = ++ 11 012 2 2 A sin right hand rule into the page d IIxdx Fk = + 2 ej ± B I 1 x I 2 h FIG. P30.52 The whole force is the sum of the forces on all of the elements of the bar: kk = + −= + = + = +− = + L N M M O Q P P = × = −− zz 0 0 0 2 7 2 2 32 2 4 2 4 4 10 200 05 10 21 0 4 0 1 1 2 01 0 II xdx hh x a fa f a f a f af ± ±± ln ± ln ln ± ln . . ± ln . ± AA A A N100 A A A cm cm cm N N P30.53 On the axis of a current loop, the magnetic field is given by B IR xR = + 0 2 2 where in this case I q = 2 πω bg . The magnetic field is directed away from the center, with a magnitude of B Rq = + = × + µω 0 2 0 2 6 2 2
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online