Chapter 19, Problem 72QAP

To determine

(a)

The direction at which the wire resting on two parallel horizontal rails would accelerate due to a magnetic field

Answer to Problem 72QAP

The force on the wire due to the magnetic force will point towards the right and hence the wire will accelerate to the right.

Explanation of Solution

Given:

A steady current of 100 A passes through a wire (of weight m=40 g=0.040 kg; length = 0.8 m) that can slide on two parallel, horizontal conducting rails. A uniform magnetic field with a magnitude of 1.2 T is directed into the page.

Formula used:

Right hand rule for the field directionIf you point your right thumb in the direction of the current and curl your fingers, the magnetic field curls around the field lines in the direction of the curled fingers of your right hand.

Calculation:

Application of the right-hand rule will give the direction of magnetic force on the wire that is initially at rest on two parallel conducting rails.

Conclusion:

The force on the wire due to the magnetic force will point towards the right and hence the wire will accelerate to the right.

To determine

(b)

The magnetic force on the wire that rests on two parallel, horizontal conducting rails.

Answer to Problem 72QAP

The magnetic force on the wire = 100 N (rounded to one significant figure)

Explanation of Solution

Given:

A steady current of 100 A passes through a wire (of weight m=40 g=0.040 kg; length = 0.8 m) that can slide on two parallel, horizontal conducting rails. A uniform magnetic field with a magnitude of 1.2 T is directed into the page.

Formula used:

The magnitude of the magnetic force acting on the wire due to a magnetic field is given by the following equation

  $\mathcal{l}\text{ = length of the wire}$. $\begin{array}{l}\overrightarrow{F_{magnetic}}=i.\mathcal{l}.B.\sin \theta \to (a)\\ \overrightarrow{F_{magnetic}}=\text{ magnetic force on the wire}\\ \text{i = current on the wire}\\ \text{B = magnitude of the magnetic field}\\ \theta =\text{ angle between the directions of the magnetic field and the magnetic force}\end{array}$

  $\mathcal{l}\text{ = length of the wire}$

Calculation:

The angle between the magnetic field and the force is 90 degrees as they are perpendicular to each other. Substituting the values to equation (a);

  $\begin{array}{l}\mathcal{l}=0.8\text{ m;B =1}\text{.2 T;}\theta {\text{=90}}^0;i=100A;\\ \overrightarrow{F_{magnetic}}=(100A)\times 0.8\text{ m}\times \text{(1}\text{.2 T)}\times {\text{sin(90}}^0)=96\text{ N=100 N(rounded to one sig}\text{.fig)}\end{array}$

Conclusion:

The magnetic force on the wire = 100 N (rounded to one significant figure)

To determine

(c)

How long must the rails be of the wire starting from rest is to reach a speed of 200 m/s.

Answer to Problem 72QAP

The length of the rails to facilitate wire reaching a speed of 200 m/s= 8 m

Explanation of Solution

Given:

A steady current of 100 A passes through a wire (of weight m=40 g=0.040 kg; length = 0.8 m) that can slide on two parallel, horizontal conducting rails. A uniform magnetic field with a magnitude of 1.2 T is directed into the page.

Also, from the part b) above we have calculated the magnitude of magnetic force on the wire to be 96 N.

Formula used:

  $\begin{array}{l}{\displaystyle \sum F_{ext,x}}=m{a}_x\to (b)\\ \text{where }{\displaystyle \sum F_{ext,x}}=\text{ total of the external forces acting on the wire along x direction}\\ m=mass\text{ of the wire}\\ {\text{a}}_x\text{ =accleration of the wire in x direction}\end{array}$

  $\begin{array}{l}{\text{v}}_x^2{\text{ =v}}_{x_0}^2{\text{ -2a}}_x(x-x_0)\to (c)\\ {\text{where v}}_x=velocity\text{ at distance x}\\ {\text{v}}_{x_0}=velocity\text{ at the starting point}\\ {\text{a}}_x\text{=accleration of the wire in x direction}\end{array}$

Calculation:

One can determine the direction of the magnetic force using the right hand rule as mentioned in part a).Assuming that the magnetic force is the only force that is acting on the wire along the horizontal direction(x direction), the wire would undergo constant acceleration towards x direction. Hence one could apply equation (b) to the motion of the wire;

  $\overrightarrow{F_{magnetic}}=m{\text{a}}_x\Rightarrow {\text{a}}_x=\frac{\overrightarrow{F_{magnetic}}}{m}$

Applying this result to (c)

  $\begin{array}{l}rearranging\text{ (c)}\\ {\text{2a}}_x(x-x_0)={\text{v}}_x^2-{\text{v}}_{x_0}^2\\ \Rightarrow (x-x_0)=\frac{{\text{v}}_x^2-{\text{v}}_{x_0}^2}{{\text{2a}}_x}\\ \Rightarrow \Delta x\text{ =}\frac{{\text{v}}_x^2-{\text{v}}_{x_0}^2}{{\text{2a}}_x}where\Delta x=(x-x_0)=lengthof\text{ the rails}\\ Substituting{\text{ v}}_x=200{\text{ m/s; v}}_{x_0}=0\text{ m/s}\\ {\text{a}}_x=\frac{\overrightarrow{{\text{F}}_{\text{magnetic}}}}{\text{m}};\overrightarrow{F_{magnetic}}=96N;m=40\text{ g=0}\text{.040 kg}\\ \Delta x=\frac{{(200m/s)}^2-{(0\text{ m/s)}}^2}{2\times \frac{96\text{ N}}{0.04\text{ kg}}}=\frac{0.04\times {(200)}^2}{192}m=8\text{ m(sig fig)}\end{array}$

Conclusion:

The length of the rails to facilitate wire reaching a speed of 200 m/s= 8 m

To determine

(d)

What would be the direction of the movement of the wire if the magnetic field was directed out of the page

Answer to Problem 72QAP

If the magnetic field is directed out of the page, force on the wire due to the magnetic field would point to the left and the wire would accelerate in that direction. The numerical value of the magnetic force would be the same as calculated in part b).

Explanation of Solution

Given:

A steady current of 100 A passes through a wire (of weight m=40 g=0.040 kg; length = 0.8 m) that can slide on two parallel, horizontal conducting rails. A uniform magnetic field with a magnitude of 1.2 T is directed out of the page.

Formula used:

Right hand rule for the field directionIf you point your right thumb in the direction of the current and curl your fingers, the magnetic field curls around the field lines in the direction of the curled fingers of your right hand

Calculation:

Application of the right hand rule would give the direction of movement of the wire.

Conclusion:

If the magnetic field is directed out of the page, force on the wire due to the magnetic field would point to the left and the wire would accelerate in that direction. The numerical value of the magnetic force would be the same as calculated in part b).

To determine

(e)

What would be the direction of the movement of the wire if the magnetic field was directed towards the top of the page

Answer to Problem 72QAP

If the magnetic field is directed towards to the top of the page the magnetic force on the wire would be zero because the current and the magnetic field would be antiparallel to each other.

Explanation of Solution

Given:

A steady current of 100 A passes through a wire (of weight m=40 g=0.040 kg; length = 0.8 m) that can slide on two parallel, horizontal conducting rails. A uniform magnetic field with a magnitude of 1.2 T is directed out of the page.

Formula used:

Right hand rule for the field directionIf you point your right thumb in the direction of the current and curl your fingers, the magnetic field curls around the field lines in the direction of the curled fingers of your right hand

Calculation:

The magnetic field and the current must not be antiparallel to facilitate a movement of the wire.

Conclusion:

If the magnetic field is directed towards to the top of the page the magnetic force on the wire would be zero because the current and the magnetic field would be antiparallel to each other