Lesson_4.2_Homework_and_Solutions

# Lesson_4.2_Homework_and_Solutions - Lesson 4.2 Homework...

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Unformatted text preview: Lesson 4.2 Homework Solutions 1. A proton traveling with a velocity of v = 1 × 10 4 ms-1 i is located at x = 3.0 m, y = 4.0 m at a given time. Find the magnetic field at (i) (2, 2), and (ii) (6, 4) at this instant. Solution (i) The vector r from the point (3m, 4m), to the point (2 m, 2 m) is (2 – 3) i + (2 – 4) j = – i – 2 j 2 2 1 2 5 r m = + = If θ is the angle between vectors r and v 1 2 tan 63.4 1 o θ- = = 1 2 9 4 1 3 7 2 2.86 1 sin 1.602 10 10 sin 63.4 10 4 5 o o qv C m k s T B r μ θ π---- × × × × = = × = (ii) The vector r from the point (3m, 4m), to the point (6 m, 4 m) is (2 – 3) i + (2 – 4) j = – 3 i – 2 j r 2 = 3 2 + 2 2 = 13 1 2 tan 33.7 3 o θ- = = 19 4 1 7 24 2 sin 1.602 10 10 sin 33.7 10 4 1 6 8 10 3 . 4 o o qv C ms B r Tk μ θ π---- × × × = = × = × 2. An electron orbits a proton at a distance of 5.29 × 10-11 m. What is the magnetic filed at the proton due to the orbital motion of the electron. Solution θ = 63 o r = - i – 2 j First we obtain the orbital velocity of the electron by using the fact that the centripetal force to keep the electron in orbit is provided by the electrostatic force between the proton and the electron. 2 2 2 mv e k r r = 9 19 6 1 11 31 9 10 1.602 10 2.2 10 5.29 10 9.1 10 k v e ms rm---- × = = × = × × × × As the electron orbits the proton, its velocity is always at right angles to the radius so that sin θ = 1 19 6 1 7 2 11 2 sin 1.602 10 2.2 10 10 4 (5.29 10 ) 12.6 o qv C ms B r T μ θ π---- × × × = = × × = 3. Two equal charges q located at (0, 0) and (0, b) are moving in the positive x-direction with equal speeds v. Find the ratio of the magnitude of the magnetic and electrostatic forces on each. Solution The distance between the two charges is b and it is at right angles to the direction of their velocities so that sin θ = 1. If B is the magnetic field produced by the motion of one charge at the other, the magnetic force F B on it is F B = qvB 2 4 o qv B b μ π = 2 2 2 4 o B q v F qvB b μ π = = The electric force F E between the two charges is given by Coulomb’s law....
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Lesson_4.2_Homework_and_Solutions - Lesson 4.2 Homework...

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