Question

A long, rigid conductor, lying along an x axis, carries a current of 4.99 A in the negative x direction. A magnetic field is present, given by = 3.72 + 8.72 x2, with x in meters and in milliteslas. Find (a) the x-component, (b) the y-component, and (c) the z-component of the force on the 1.36 m segment of the conductor that lies between x = 1.41 m and x = 2.77 m.

Answer 1

Answer with Explanation:

We are given that

Current in conductor=I=4.99 A  (-x direction)

Magnetic field=B=$3.72\hat{i}+8.72x^2\hat{j}mT=(3.72i+8.72x^2j)\times 10^{-3}$

(1mT=$10^{-3} T$)

x(in m) and B (in mT)

Length of conductor is given in negative x- direction

$\vec{L}=-x\hat{i}$

$dL=-dx\hat{i}$

Force on current carrying conductor is given by

$F=I(L\times B)$

$dF=I(dL\times B)$

Integrating on both sides then we get

$\vec{F}=\int_{1.41}^{2.77}(4.99)(-dx\hat{i}\times (3.72\hat{i}+8.72x^2\hat{j}))\times 10^{-3}$

$\vec{F}=-\int_{1.41}^{2.77}(4.99\times 10^{-3})\cdot 8.72(x^2\hat{k})dx$  ($i\times i=0, i\times j=k$

$\vec{F}=-(4.99\times 10^{-3}\times 8.72)[\frac{x^3\hat{k}}{3}]^{2.77}_{1.41}$

$\vec{F}=-\frac{(4.99\times 10^{-3}\cdot 8.72)}{3}((2.77)^3-(1.41)^3)\hat{k}$

$\vec{F}=-0.268 \hat{k} N$

a. x- component of force=0

b.y- component of force=0

c.z- component of force=-0.268 N