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(14.1) A current -3A flows in a straight wire with length L 6 m directed along t

ID: 1793944 • Letter: #

Question

(14.1) A current -3A flows in a straight wire with length L 6 m directed along the -axis. Our goal is to determine the approximate value of the magnetic field produced by the current in the wire at a field point located 1m from the 2/3 point of the wire. 3 (a) Using the Law of Biot and Savart, compute the magnitude and direction of the mag- netic field produced by the current in the wire at the location of the field point. (b) Modelling the wire as a single current element, IAs = IoL = 10 Li, located at the centre of the wire (the origin), approximate the magnitude and direction of the magnetic field at P (c) Modelling the wire as two current elements, 1G-10L/2 = lo, located sym- metrically about the centre of the wire (at x = ±3/2), approximate the magnitude and direction of the magnetic field at P. (d) Modelling the wire as three current elements, I as-10 L/3-10 , located symmet- rically about the centre of the wire (at-2,0,,2), approximate the magnitude and tion of the magnetic field at P. (e) Modelling the wire as four current elements Is-6 L/4 = 10 4t, located symmetri- cally about the centre of the wire (x =-9/4,-3/4, 3/4,9/4), approximate the magnitude and direction of the magnetic field at P

Explanation / Answer

14.1 Io = 3 A

L = 6 m

d = 1m

a. from biot savart's law we know magnetic field of a finite wire at distance d is

B(d) = kI(cos(alpha) + cos(beta))/d

from the given question

cos(theta) = 4/5

cos(beta) = 2/sqroot(5)

hence

B(1m) = 10^-7*3(4/5 + 2/sqrt(5))/1

B = 5.08328*10^-7 T (out of the page)

b. considering this wire as a single element

from biot savarts law

dB = k*I x dl/r^2

P = (1, 1)

r = sqroot(2)

hence

B = 10^-7*3sin(45)/2 = 1.0606*10^-7 T (out of the page)

c. modelling them as two wire segments

from biot savarts law

P = (1,1)

origins = (1.5,0),(-1.5,0)

B = kI*1/(0.5^2 + 1)^3/2 + kI*1/(2.5^2 + 1)^3/2 = 2.3*10^-7 T (out of the page)

d. modelling as three elements

origins = (-2,0),(0,0),(2,0)

B = kI ( 1/(3^2 + 1)^3/2 + 1/(1^2 + 1^2)^3/2 + 1/(1^2 + 1)^3/2)

B = 2.216*10^-7 T