Explanation 1) Concept: As the projectile is moving in the presence of magnetic field due to the two rails (1 and 2), it is also carrying current i . So small current element of projectile will experience force due to both rails. Net force on current element ( i d l → ) will be the vector sum of individual forces due to rail 1 and 2. 2) Formula: i) Force acing on current element kept in magnetic field, d F → = i d l → × B → ii) Magnetic field due to long wire carrying current i , at distance r is B = μ 0 i 2 π r 3) Given: i) Radius of circular cross-section of the rail, R = 6.7 c m = 6.7 × 10 - 2 m ii) Distance between the rails, w = 12.0 m m = 12 × 10 - 3 m iii) Length of the rail, L = 4.0 m iv) Current through the rail, i = 450 × 10 3 A v) Mass of the projectile is m = 10 g = 10 × 10 - 3 k g 4) Calculations: a) Consider a portion of projectile between y & y + d y . We can find magnetic force on this segment due to both rails (1 and 2) The net force on the segment is F → = d F 1 → + d F 2 → , where d F 1 → and d F 2 → are the forces on portion of projectile due to rail 1 and 2 respectively. d F → = d F 1 → + d F 2 → = i d l → × B 1 → + i d l → × B 2 → We can find B field due to rail 1 at distance ( 2 R + w - y ) from the center of rail 1. We can use Ampere law for this, then we get B 1 = μ 0 i 2 π 2 R + w - y ( - k ^ ) Similarly, B field due to rail 2 at distance ( y ) from the center of rail 2, is B 2 = μ 0 i 2 π y ( - k ^ ) And d l → = - d y · j ^ using this we get d F → = d F 1 → + d F 2 → d F → = i d y · ( - j ^ ) × μ 0 i 2 π 2 R + w - y ( - k ^ ) + i d y · ( - j ^ ) × μ 0 i 2 π y ( - k ^ ) By using cross product rule j ^ × k ^ = i ^ d F → = i · i ^ μ 0 i d y 2 π 2 R + w - y + i ^ μ 0 i d y 2 π y d F → = i d y · μ 0 i 2 π 2 R + w - y + μ 0 i 2 π y i ^ Total force is obtained by integrating d F → over

FUND. OF PHYSICS FOR LSU WILEY+ NEXT GEN

11th Edition

ISBN: 9781119749295

Author: Halliday

Publisher: WILEY

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Magnetic Force

A magnetic force arises when a charged particle or a current-carrying conductor is placed in a magnetic field. It is created because of the movement of the charged particles.

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Chapter 29, Problem 88P

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a) Force on the projectile

b) Speed of the projectile

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Chapter 29, Problem 87P

Chapter 30, Problem 1Q

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Chapter 29 Solutions

FUND. OF PHYSICS FOR LSU WILEY+ NEXT GEN

Ch. 29 - Prob. 1Q Ch. 29 - Prob. 2Q Ch. 29 - Prob. 3Q Ch. 29 - Prob. 4Q Ch. 29 - Prob. 5Q Ch. 29 - Prob. 6Q Ch. 29 - Prob. 7Q Ch. 29 - Prob. 8Q Ch. 29 - Prob. 9Q Ch. 29 - Prob. 10Q

Ch. 29 - Prob. 11Q Ch. 29 - A surveyor is using a magnetic compass 6.1 m below...Ch. 29 - Figure 29-35a shows an element of length ds = 1.00...Ch. 29 - SSM At a certain location in the Philippines,...Ch. 29 - Prob. 4P Ch. 29 - Prob. 5P Ch. 29 - Prob. 6P Ch. 29 - Prob. 7P Ch. 29 - Prob. 8P Ch. 29 - Prob. 9P Ch. 29 - Prob. 10P Ch. 29 - Prob. 11P Ch. 29 - Prob. 12P Ch. 29 - Prob. 13P Ch. 29 - Equation 29-4 gives the magnitude B of the...Ch. 29 - Prob. 15P Ch. 29 - Prob. 16P Ch. 29 - Prob. 17P Ch. 29 - Prob. 18P Ch. 29 - Prob. 19P Ch. 29 - Prob. 20P Ch. 29 - Prob. 21P Ch. 29 - Prob. 22P Ch. 29 - Prob. 23P Ch. 29 - Prob. 24P Ch. 29 - Prob. 25P Ch. 29 - Prob. 26P Ch. 29 - Prob. 27P Ch. 29 - GO Figure 29.56a shows two wires, each carrying....Ch. 29 - Prob. 29P Ch. 29 - Prob. 30P Ch. 29 - Prob. 31P Ch. 29 - GO The current-carrying wire loop in Fig. 29-6a...Ch. 29 - Prob. 33P Ch. 29 - Prob. 34P Ch. 29 - Prob. 35P Ch. 29 - Prob. 36P Ch. 29 - Prob. 37P Ch. 29 - Prob. 38P Ch. 29 - Prob. 39P Ch. 29 - Prob. 40P Ch. 29 - Prob. 41P Ch. 29 - Prob. 42P Ch. 29 - Prob. 43P Ch. 29 - Prob. 44P Ch. 29 - Prob. 45P Ch. 29 - Prob. 46P Ch. 29 - ILW The current density inside a long, solid,...Ch. 29 - Prob. 48P Ch. 29 - Prob. 49P Ch. 29 - A solenoid that is 95.0 cm long has a radius of...Ch. 29 - A 200-turn solenoid having a length of 25 cm and a...Ch. 29 - A solenoid 1.30 m long and 2.60 cm in diameter...Ch. 29 - A long solenoid has 100 turns/cm and carries...Ch. 29 - An electron is shot into one end of a solenoid. As...Ch. 29 - Prob. 55P Ch. 29 - Prob. 56P Ch. 29 - Prob. 57P Ch. 29 - Prob. 58P Ch. 29 - Prob. 59P Ch. 29 - Prob. 60P Ch. 29 - A circular loop of radius 12 cm carries a current...Ch. 29 - Prob. 62P Ch. 29 - Prob. 63P Ch. 29 - Prob. 64P Ch. 29 - A cylindrical cable of radius 8.00 mm carries a...Ch. 29 - Two long wires lie in an xy plane, and each...Ch. 29 - Two wires, both of length L, are formed into a...Ch. 29 - Prob. 68P Ch. 29 - Prob. 69P Ch. 29 - Prob. 70P Ch. 29 - A 10-gauge bare copper wire 2.6 mm in diameter can...Ch. 29 - A long vertical wire carries an unknown current....Ch. 29 - Prob. 73P Ch. 29 - The magnitude of the magnetic field at a point...Ch. 29 - Prob. 75P Ch. 29 - Prob. 76P Ch. 29 - Prob. 77P Ch. 29 - A long wire carrying 100 A is perpendicular to the...Ch. 29 - A long, hollow, cylindrical conductor with inner...Ch. 29 - A long wire is known to have a radius greater than...Ch. 29 - Prob. 81P Ch. 29 - Prob. 82P Ch. 29 - Prob. 83P Ch. 29 - Three long wires all lie in an xy plane parallel...Ch. 29 - Prob. 85P Ch. 29 - Prob. 86P Ch. 29 - Prob. 87P Ch. 29 - Prob. 88P

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a) Force on the projectile b) Speed of the projectile

Solution Summary: The author explains that the projectile is moving in the presence of magnetic field due to the two rails (1 and 2), it is also carrying current i.

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To find: a) Force on the projectile b) Speed of the projectile

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Chapter 29 Solutions

To find:

a) Force on the projectile

b) Speed of the projectile