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)* 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* B)`

`dF=I(dL* B)`

Integrating on both sides then we get

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

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

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

`\vec{F}=-((4.99* 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