PHYSICS F/ SCI +ENGRS W/ WEBASSIGN ACCES
PHYSICS F/ SCI +ENGRS W/ WEBASSIGN ACCES
10th Edition
ISBN: 9781337888509
Author: SERWAY
Publisher: CENGAGE L
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Chapter 29, Problem 42AP

Review. Rail guns have been suggested for launching projectiles into space without chemical rockets. A tabletop model rail gun (Fig. P29.42) consists of two long, parallel, horizontal rails = 3.50 cm apart, bridged by a bar of mass m = 3.00 g that is free to slide without friction. The rails and bar have low electric resistance, and the current is limited to a constant I = 24.0 A by a power supply that is far to the left of the figure, so it has no magnetic effect on the bar. Figure P29.42 shows the bar at rest at the midpoint of the rails at the moment the current is established. We wish to find the speed with which the bar leaves the rails after being released from the midpoint of the rails. (a) Find the magnitude of the magnetic field at a distance of 1.75 cm from a single long wire carrying a current of 2.40 A. (b) For purposes of evaluating the magnetic field, model the rails as infinitely long. Using the result of part (a), find the magnitude and direction of the magnetic field at the midpoint of the bar. (c) Argue that this value of the field will be the same at all positions of the bar to the right of the midpoint of the rails. At other points along the bar, the field is in the same direction as at the midpoint, but is larger in magnitude. Assume the average effective magnetic field along the bar is five times larger than the field at the midpoint. With this assumption, find (d) the magnitude and (e) the direction of the force on the bar. (f) Is the bar properly modeled as a particle under constant acceleration? (g) Find the velocity of the bar after it has traveled a distance d = 130 cm to the end of the rails.

Figure P29.42

Chapter 29, Problem 42AP, Review. Rail guns have been suggested for launching projectiles into space without chemical rockets.

(a)

Expert Solution
Check Mark
To determine
The magnitude of the magnetic field at a distance of 1.75cm from a single long wire.

Answer to Problem 42AP

The magnitude of the magnetic field at a distance of 1.75cm from a single long wire is 2.74×104T .

Explanation of Solution

Given info: The length of the rails is 3.50cm , the mass of bar is 3.00g , value of current is 24.0A .

Formula to calculate the magnetic field is

B=μ0I2πr (1)

Here,

B is the magnetic field.

μ0 is the absolute permeability.

I is the current flow in the wire.

r is the distance of the point.

Substitute 4π×107Tm/A for μ0 , 24.0A for I , and 1.75cm for r to find B .

B=(4π×107Tm/A)×24.0A2π×1.75cm×102m1cm=2.74×104T

Conclusion:

Therefore, the magnitude of the magnetic field at a distance of 1.75cm from a single long wire is 2.74×104T .

(b)

Expert Solution
Check Mark
To determine
The magnitude and direction of the magnetic field at the mid-point of the bar.

Answer to Problem 42AP

The magnitude of the magnetic field at the mid-point of the bar is 2.74×104T into the page.

Explanation of Solution

Given info: The length of the rails is 3.50cm , the mass of bar is 3.00g , value of current is 24.0A .

The diagram that represents the given situation is shown below.

PHYSICS F/ SCI +ENGRS W/ WEBASSIGN ACCES, Chapter 29, Problem 42AP

Figure (I)

Since the current is diverted through the bar, only half of each rails carries currents, so the field produce by each rail is half of the infinitely long wire produces.

Write the expression for the magnetic field produce by conductor AB at point C is,

BAB=12B

Substitute 2.74×104T for B in above equation.

BAB=12×2.74×104T=1.37104T

Write the expression for the magnetic field produce by conductor DE at point C is,

BDE=12B

Substitute 2.74×104T for B in above equation.

BDE=12×2.74×104T=1.37104T

Formula to calculate the total magnetic field at point C is,

BC=BAB+BDE

Substitute 1.37104T for BAB and BDE in above equation.

BC=1.37104T+1.37104T=2.74×104T

Conclusion:

Therefore, the magnitude of the magnetic field at the mid-point of the bar is 2.74×104T into the page.

(c)

Expert Solution
Check Mark
To determine
The reason that the value of magnetic field will be same at all position of the bar to the right of the midpoint of the rails.

Answer to Problem 42AP

The rails are very long so the location of the bar does not depend upon the length of the rail to the right side.

Explanation of Solution

Given info: The length of the rails is 3.50cm , the mass of bar is 3.00g , value of current is 24.0A .

The assumption makes that the rails are infinitely long so, the length of the rail to the right of the bar does not depend upon the location of the bar. Hence the magnetic field will be same at all position of the bar to the right of the midpoint of the rails.

Conclusion:

Therefore, due to infinite length of the rails makes the field same at all position of the midpoint of the bar.

(d)

Expert Solution
Check Mark
To determine
The magnitude of the force on the bar.

Answer to Problem 42AP

The magnitude of the force on the bar is 1.15×103N .

Explanation of Solution

Given info: The length of the rails is 3.50cm , the mass of bar is 3.00g , value of current is 24.0A .

Formula to calculate the force on the bar is,

F=I(l×B)

Here,

I is the value of current.

l is the length vector.

B is the field vector.

Substitute 24.0A for I , 3.50cmk^ for l and (5×2.74×104T) for B to find F

F=(24.0A)(3.50cm×102m1cmk^)(5×2.74×104T)=1.15×103Ni^

Conclusion:

Therefore, the magnitude of the force on the bar is 1.15×103N .

(e)

Expert Solution
Check Mark
To determine
The direction of the force on the bar.

Answer to Problem 42AP

The direction of the force on the bar is in positive x-direction.

Explanation of Solution

Given info: The length of the rails is 3.50cm , the mass of bar is 3.00g , value of current is 24.0A .

The force vector on the bar is 1.15×103Ni^ . So, the direction of the force on the bar is in the positive x-direction.

Conclusion:

Therefore, the direction of the force on the bar is in positive x-direction.

(f)

Expert Solution
Check Mark
To determine
Whether the bar is properly modeled as a particle under constant acceleration.

Answer to Problem 42AP

Yes, the bar will move with constant acceleration of magnitude 0.384m/s2 .

Explanation of Solution

Given info: The length of the rails is 3.50cm , the mass of bar is 3.00g , value of current is 24.0A .

The length of the bar, value of current and field producer is constant so, the force exerted on the bar is constant that gives uniform acceleration of the bar.

Formula to calculate the acceleration of the bar is,

a=Fm

Substitute 1.15×103N for F and 3.00g for m to find a .

a=1.15×103N3.00g×103kg1g=0.384m/s2

Conclusion:

Therefore, the bar will move with constant acceleration of magnitude 0.384m/s2 .

(g)

Expert Solution
Check Mark
To determine
The velocity of the bar.

Answer to Problem 42AP

The velocity of the bar is 0.998m/s .

Explanation of Solution

Given info: The length of the rails is 3.50cm , the mass of bar is 3.00g , value of current is 24.0A .

Formula to calculate the velocity of the bar is,

vf2=vi2+2ad

Here,

vf is the final velocity of the bar.

vi is the initial velocity of the bar.

d is the distance travelled by the bar.

Substitute 0 for vi , 0.384m/s2 for a and 130cm for d to find vf .

vf=0+2(0.384m/s2)(130cm×102m1cm)=0.998m/s

Conclusion:

Therefore, the velocity of the bar is 0.998m/s .

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Chapter 29 Solutions

PHYSICS F/ SCI +ENGRS W/ WEBASSIGN ACCES

Ch. 29 - Consider a flat, circular current loop of radius R...Ch. 29 - Prob. 7PCh. 29 - One long wire carries current 30.0 A to the left...Ch. 29 - Determine the magnetic field (in terms of I, a,...Ch. 29 - Prob. 10PCh. 29 - Two long, parallel wires carry currents of I1 =...Ch. 29 - Prob. 12PCh. 29 - Prob. 13PCh. 29 - Prob. 14PCh. 29 - You are part of a team working in a machine parts...Ch. 29 - Why is the following situation impossible? Two...Ch. 29 - Prob. 17PCh. 29 - Niobium metal becomes a superconductor when cooled...Ch. 29 - The magnetic coils of a tokamak fusion reactor are...Ch. 29 - A packed bundle of 100 long, straight, insulated...Ch. 29 - The magnetic field 40.0 cm away from a long,...Ch. 29 - Prob. 22PCh. 29 - A long solenoid that has 1 000 turns uniformly...Ch. 29 - A certain superconducting magnet in the form of a...Ch. 29 - Prob. 25PCh. 29 - You are given a certain volume of copper from...Ch. 29 - Prob. 27PCh. 29 - You are working for a company that creates special...Ch. 29 - A solenoid of radius r = 1.25 cm and length =...Ch. 29 - The magnetic moment of the Earth is approximately...Ch. 29 - A 30.0-turn solenoid of length 6.00 cm produces a...Ch. 29 - Why is the following situation impossible? The...Ch. 29 - Suppose you install a compass on the center of a...Ch. 29 - Prob. 34APCh. 29 - A nonconducting ring of radius 10.0 cm is...Ch. 29 - Prob. 36APCh. 29 - A very large parallel-plate capacitor has uniform...Ch. 29 - Two circular coils of radius R, each with N turns,...Ch. 29 - Prob. 39APCh. 29 - Two circular loops are parallel, coaxial, and...Ch. 29 - As seen in previous chapters, any object with...Ch. 29 - Review. Rail guns have been suggested for...Ch. 29 - Prob. 43APCh. 29 - An infinitely long, straight wire carrying a...Ch. 29 - Prob. 45CPCh. 29 - We have seen that a long solenoid produces a...Ch. 29 - A wire carrying a current I is bent into the shape...Ch. 29 - Prob. 48CPCh. 29 - Prob. 49CPCh. 29 - Prob. 50CPCh. 29 - The magnitude of the force on a magnetic dipole ...
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