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7C%20S03%20Bk%2013%20Exercises%20and%20Q-Ps

# 7C%20S03%20Bk%2013%20Exercises%20and%20Q-Ps - Physics 7C...

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Physics 7C Spring 2003: Block 13 Exercises and Q/Ps 1 03 03 20 Examples 1 .The spacecraft Apollo 13 (of mass 10,000 kg) is 90% of the way from its journey from the Earth to the Moon. (a)Determine the total gravitational field due to both the Earth and the Moon at this location. Draw these individual field vectors, as well as the total field vector. (b)Determine the net force F on the Apollo 13 spacecraft. Solution (a)The distance from the Earth to the Moon is 3.85 × 10 8 m. The magnitude of the gravitational field due to the Earth at 90% of this distance is: g Earth = G M Earth r 2 = 6.672 × 10 11 N m 2 kg 2 5.975 × 10 24 kg 9 10 3.85 × 10 8 m 2 = 3.32 × 10 3 N kg . The magnitude of the gravitational field due to the Moon at this distance is: g Moon = G M Moon r 2 = 6.672 × 10 11 N m 2 kg 2 7.35 × 10 22 kg 1 10 3.85 × 10 8 m 2 = 2.68 × 10 3 N kg . The magnitude of the total gravitational field vector is then the tail-to-head vector addition of the individual gravitational field vectors. g 3.85 × 10 m 8 Earth Moon Apollo 13 g Earth g Moon g Earth g Moon Apollo 13 = g Earth + g Moon g So the magnitude of the total gravitational field vector is then: Physics 7C Spring 2003: Block 13 Exercises and Q/Ps 2 03 03 20 g = g Earth + g Moon = 3.32 × 10 3 N kg + 2.68 × 10 3 N kg = − 6. 4 × 10 4 N kg , where the negative sign tells us that the total gravitational field vector points back towards the direction of Earth. (b)Since the only forces that act on the Apollo 13 spacecraft are gravitational forces, then the net force F is equal to the total gravitational force, which points back towards the Earth. The magnitude of the net force F on the Apollo 13 spacecraft is then: F on Apollo 13 = m Apollo 13 g = 10, 000 kg ( ) 6.4 × 10 4 N kg = 6.4 N. 2 .Calculate the magnitude and direction of the magnetic field due to a wire carrying 1 Amp of current into the plane of this page, and a wire carrying 1 Amp of current out of the plane of this page, at the location x shown at left. Solution The directions of the magnetic fields due to current 1 and current 2 are given by RHR1—we put our thumb into the page for current 1, such that the magnetic field at location x is directed down in the plane of this page. We put our thumb out of the page for current 2, such that the magnetic field at location x is directed up along the plane of this page. The magnitudes of the magnetic field vectors at this location are not the same, however, so these magnetic field vectors will not totally c ancel each other. The magnitude of the magnetic field due to current 1 is: B 1 = µ 0 ( ) I 1 2 π r 1 = 1.26 × 10 6 Tesla m Amps 1 Amp 2 π 0.2 m ( ) = 1.00 × 10 6 Teslas , where r 1 x . And similarly for t he magnitude of the magnetic field due to current 2: B 2 = µ 0 ( ) I 2 2 π r 2 = 1.26 × 10 6 Tesla m Amps 1 Amp 2

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