Question

Three infinitely long, straight wires with distance d are placed in a row. The current flowing through each wire is I. Assuming that the radius of the wires is negligible (a) Find two spatial locations at which the magnetic field is zero. (b) Plot the magnetic field lines. (c) Supposing that the current in the middle wire changes to pposite direction, namely downwards, and then the middle wire moves an infinitesimal displacement (d), and other wires stay still. Find the dynamic equation for the middle wire. (Note: the dynamic equation is used to describe the relationship between the net force on the middle wire and the displacement ax), and identify qualitatively the motion in terms of the dynamic equation you have found

please kindly help me

Answer 1

$\textrm{(a)}$

$\textrm{Let us consider the region between wire 1 and wire 2.}$

$\textrm{Consider a point, which is at a distance of } \ r\textrm{ from wire 1.}$

$\textrm{This point is at a distance of } \ d-r \textrm{ from wire 2.}$

$\textrm{Also it is at a distance of } \ 2d-r \textrm{ from wire 3.}$

$\textrm{The magnetic field at the point } r \textrm{ due to wire 1 is }B_1\textrm{ which is given by}$

$\vec{B}_1 \ = \ - \ \frac{\mu_0 \ I}{2\pi r} \ \hat{k}$

$\textrm{This field in the plane of the paper is penetrating into the page.}$

$\textrm{The magnetic field at the point } r \textrm{ due to wire 2 is }B_2\textrm{ which is given by}$

$\vec{B}_2 \ = \ \frac{\mu_0 \ I}{2\pi (d-r)} \ \hat{k}$

$\textrm{This field in the plane of the paper is penetrating out of the page.}$

$\textrm{The magnetic field at the point } r \textrm{ due to wire 3 is }B_3\textrm{ which is given by}$

$\vec{B}_2 \ = \ \frac{\mu_0 \ I}{2\pi (2d-r)} \ \hat{k}$

$\textrm{This field in the plane of the paper is penetrating out of the page.}$

$\textrm{Now the total field at the point }r\textrm{ is }$

$\vec{B} \ = \ \vec{B}_1 \ + \ \vec{B}_2 \ + \ \vec{B}_3$

$\Rightarrow \ \ \vec{B} \ = \ - \ \frac{\mu_0 \ I}{2\pi r} \ \hat{k} \ + \ \frac{\mu_0 \ I}{2\pi (d-r)} \ \hat{k} \ + \ \frac{\mu_0 \ I}{2\pi (2d-r)} \ \hat{k}$

$\Rightarrow \ \ \vec{B} \ = \ \frac{\mu_0 \ I}{2\pi r} \ \left ( -\frac{1}{r} \ + \ \frac{1}{d-r} \ + \ \frac{1}{2d-r} \right ) \ \hat{k}$

$\textrm{For }\vec{B} = 0 \ \textrm{ we must have }\ \left ( -\frac{1}{r} \ + \ \frac{1}{d-r} \ + \ \frac{1}{2d-r} \right ) \ = \ 0$

$\Rightarrow \ \ -\frac{1}{r} \ + \ \frac{1}{d-r} \ + \ \frac{1}{2d-r} \ = \ 0$

$\Rightarrow \ \ \frac{- (d-r) \ (2d-r) \ + \ r \ (2d-r) \ + \ r \ (d-r) }{r \ (d-r) \ (2d-r)} \ = \ 0$

$\Rightarrow \ \ - (d-r) \ (2d-r) \ + \ r \ (2d-r) \ + \ r \ (d-r) \ = \ 0$

$\Rightarrow \ \ - \ (2d^2 + r^2 - dr - 2dr) \ + \ (2dr - r^2) \ + \ (dr - r^2) \ = \ 0$

$\Rightarrow \ \ - \ 3r^2 \ + \ 6dr \ - \ 2d^2 \ = \ 0$

$\Rightarrow \ \ 3r^2 \ - \ 6dr \ + \ 2d^2 \ = \ 0$

$\Rightarrow \ \ r \ = \ \frac{6d \ \pm \ \sqrt{36d^2 \ - 24d^2}}{6}$

$\Rightarrow \ \ r \ = \ \frac{6d \ \pm \ \sqrt{12d^2 }}{6}$

$\Rightarrow \ \ r \ = \ \left (\frac{6 \ \pm \ 2\sqrt{3 }}{6} \right ) \ d$

$\Rightarrow \ \ r \ = \ \left (\frac{3 \ \pm \ \sqrt{3 }}{3} \right ) \ d$

$\Rightarrow \ \ r \ = \ \left (\frac{3 \ - \ \sqrt{3 }}{3} \right ) \ d \ , \ \ \left (\frac{3 \ + \ \sqrt{3 }}{3} \right ) \ d$

$\textrm{Of the above two solutions, }$

$\textrm{the first solution,} \ i.e., r = \left (\frac{3 \ - \ \sqrt{3 }}{3} \right ) d \ \textrm{ corresponds to the region between wire 1 and wire 2}$

$\textrm{the second solution,} \ i.e., r = \left (\frac{3 \ + \ \sqrt{3 }}{3} \right ) d \ \textrm{ corresponds to the region between wire 2 and wire 3}$

$\textrm{At these two points the magnetic field is zero.}$

$\textrm{(c)}$

$\textrm{Let us consider that the middle wire is displaced by }x\textrm{ towards left}, i.e.\textrm{, towards wire 1 }$

$\textrm{So the middle wire is at a distance of }(d-x)\textrm{ from wire 1 and at a distance of }$

$(d+x)\textrm{ from wire 2.}$

$\textrm{Now the magnetic field due to the current in wire 1 at a distance of }\left (d-x \right ) \ \textrm{ is given by}$

$\vec{B}_1 \ = \ - \ \frac{\mu_0 \ I}{2\pi (d-x)} \ \hat{k}$

$\textrm{Now the force per unit length due to this magnetic field on the middle wire is given by}$

$\vec{F}_1 \ = \ \left (- \ I \ \hat{j} \right ) \times \left ( - \ \frac{\mu_0 \ I}{2\pi (d-x)} \ \hat{k} \right )$

$\vec{F}_1 \ = \ \frac{\mu_0 \ I^2}{2\pi (d-x)} \ \hat{i}$

$\textrm{That is the force on the middle wire due to the current in wire 1 is towards right.}$

$\textrm{Similarly, the magnetic field due to the current in wire 3 at a distance of }\left (d+x \right ) \ \textrm{ is given by}$

$\vec{B}_3 \ = \ \frac{\mu_0 \ I}{2\pi (d+x)} \ \hat{k}$

$\textrm{Now the force per unit length due to this magnetic field on the middle wire is given by}$

$\vec{F}_2 \ = \ \left (- \ I \ \hat{j} \right ) \times \left ( \frac{\mu_0 \ I}{2\pi (d+x)} \ \hat{k} \right )$

$\vec{F}_2 \ = \ - \ \frac{\mu_0 \ I^2}{2\pi (d+x)} \ \hat{i}$

$\textrm{That is the force on the middle wire due to the current in wire 2 is towards left.}$

$\textrm{The net force per unit length on the middle wire is given by}$

$\vec{F} \ = \ \vec{F}_1 \ + \ \vec{F}_2$

$\Rightarrow \ \ \vec{F} \ = \ \frac{\mu_0 \ I^2}{2\pi (d-x)} \ \hat{i} \ - \ \frac{\mu_0 \ I^2}{2\pi (d+x)} \ \hat{i}$

$\Rightarrow \ \ \vec{F} \ = \ \frac{\mu_0 \ I^2}{2\pi } \ \left (\frac{1}{(d-x)} \ - \ \frac{1}{(d+x)} \right ) \ \ \hat{i}$

$\Rightarrow \ \ \vec{F} \ = \ \frac{\mu_0 \ I^2}{2\pi } \ \left (\frac{(d+x) \ - \ (d-x)}{(d-x) \ (d+x)} \right ) \ \ \hat{i}$

$\Rightarrow \ \ \vec{F} \ = \ \frac{\mu_0 \ I^2}{2\pi } \ \left (\frac{2x}{d^2-x^2} \right ) \ \ \hat{i}$

$\Rightarrow \ \ \vec{F} \ = \ \frac{\mu_0 \ I^2}{2\pi \ d^2 } \ \left (\frac{2x}{1-\left (x/d \right )^2} \right ) \ \ \hat{i}$

$\Rightarrow \ \ \vec{F} \ = \ \frac{\mu_0 \ I^2}{2\pi \ d^2 } \ \left \{ 2x \ \left [\frac{1}{1-\left (x/d \right )^2} \right ] \right \} \ \ \hat{i}$

$\Rightarrow \ \ \vec{F} \ = \ \frac{\mu_0 \ I^2}{2\pi \ d^2 } \ \left \{ 2x \ \left [1 \ + \ \left (\frac{x}{d} \right )^2 \ + \ . \ . \ . \ \right ] \right \} \ \ \hat{i}$

$\Rightarrow \ \ \vec{F} \ = \ \left (\frac{\mu_0 \ I^2}{\pi \ d^2 } \ x \right ) \ \hat{i} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left ( \because x<<d \right )$

$\textrm{The net force acting on the middle wire pushes it towards right.}$

$\textrm{For small displacement }x\textrm{, the force on the middle wire is proportional to }x.$