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# E2HWCH27 - Chapter 27 Sources of the Magnetic Field 13[SSM...

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Chapter 27 Sources of the Magnetic Field 13 [SSM] At time t = 0, a particle has a charge of 12 μ C, is located in the z = 0 plane at x = 0, y = 2.0 m, and has a velocity equal to i ˆ m/s 30 . Find the magnetic field in the z = 0 plane at ( a ) the origin, ( b ) x = 0, y = 1.0 m, ( c ) x = 0, y = 3.0 m, and ( d ) x = 0, y = 4.0 m. Picture the Problem We can substitute for v and q in the equation describing the magnetic field of the moving charged particle ( 2 0 ˆ 4 r q r v B × = π μ ), evaluate r and r ˆ for each of the given points of interest, and then find B . The magnetic field of the moving charged particle is given by: 2 0 ˆ 4 r q r v B × = π μ Substitute numerical values and simplify to obtain: ( 29 ( 29 ( 29 ( 29 2 2 2 2 7 ˆ ˆ m pT 0 . 36 ˆ ˆ m/s 30 C 12 N/A 10 r r r i r i B × = × = - μ ( a ) Find r and r ˆ for the particle at (0, 2.0 m) and the point of interest at the origin: ( 29 j r ˆ m 0 . 2 - = , m 0 . 2 = r , and j r ˆ ˆ - = Evaluating ( 29 0 , 0 B yields: ( 29 ( 29 ( 29 ( 29 ( 29 k j i B ˆ pT 0 . 9 m 0 . 2 ˆ ˆ m pT 0 . 36 0 , 0 2 2 - = - × = ( b ) Find r and r ˆ for the particle at (0, 2.0 m) and the point of interest at (0, 1.0 m): ( 29 j r ˆ m 0 . 1 - = , m 0 . 1 = r , and j r ˆ ˆ - = Evaluate ( 29 m 0 . 1 , 0 B to obtain: ( 29 ( 29 ( 29 ( 29 ( 29 k j i B ˆ pT 36 m 0 . 1 ˆ ˆ m pT 0 . 36 m 0 . 1 , 0 2 2 - = - × = ( c ) Find r and r ˆ for the particle at (0, 2.0 m) and the point of interest at (0, 3.0 m): ( 29 j r ˆ m 0 . 1 = , m 0 . 1 = r , and j r ˆ ˆ =

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Evaluating ( 29 m 0 . 3 , 0 B yields: ( 29 ( 29 ( 29 ( 29 k j i B ˆ pT 36 m 0 . 1 ˆ ˆ m pT 0 . 36 m 0 . 3 , 0 2 2 = × = ( d ) Find r and r ˆ for the particle at (0, 2.0 m) and the point of interest at (0, 4.0 m): ( 29 j r ˆ m 0 . 2 = , m 0 . 2 = r , and j r ˆ ˆ = Evaluate ( 29 m 0 . 4 , 0 B to obtain: ( 29 ( 29 ( 29 ( 29 k j i B ˆ pT 0 . 9 m 0 . 2 ˆ ˆ m pT 0 . 36 m 0 . 4 , 0 2 2 = × = 16 •• In a pre–quantum-mechanical model of the hydrogen atom, an electron orbits a proton at a radius of 5.29 × 10 –11 m. According to this model, what is the magnitude of the magnetic field at the proton due to the orbital motion of the electron? Neglect any motion of the proton. Picture the Problem The centripetal force acting on the orbiting electron is the Coulomb force between the electron and the proton. We can apply Newton’s 2 nd law to the electron to find its orbital speed and then use the expression for the magnetic field of a moving charge to find B . Express the magnetic field due to the motion of the electron: 2 0 4 r ev B π μ = Apply = c radial ma F to the electron: r v m r ke 2 2 2 = mr ke v 2 = Substitute for v in the expression for B and simplify to obtain: mr k r e mr ke r e B 2 2 0 2 2 0 4 4 π μ π μ = =
Substitute numerical values and evaluate B : ( 29 ( 29 ( 29 ( 29 ( 29 T 5 . 12 m 10 29 . 5 kg 10 109 . 9 C / m N 10 988 . 8 m 10 29 . 5 4 C 10 602 . 1 N/A 10 4 11 31 2 2 9 2 11 2 19 2 7 = × × × × × × = - - - - - π π B 21 A single conducting loop has a radius equal to 3.0 cm and carries a current equal to 2.6 A. What is

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E2HWCH27 - Chapter 27 Sources of the Magnetic Field 13[SSM...

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