ps08sol

ps08sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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PS08 Solutions-1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2008 Problem Set 8 Solutions Problem 1: Charges in a Uniform Field An electron with velocity ˆˆ xy vv + i j moves through a uniform magnetic field B B + i j (a) Find the magnitude of the force on the electron. (b) Repeat your calculation for a proton having the same velocity. This is just a question of calculating a cross-product. Although it wasn’t introduced in class, the way to do this is with a determinant: () ˆ ˆ 0 0 x y y x qq v v q v B v B BB = = ij k Fv B k GG G So the magnitude of the force is ( ) yx Fe v B v B =− for both an electron and proton, the only difference is the direction is reversed ( q =- e for an electron). Problem 2: Power Line A horizontal power line carries a current of I from south to north. Earth's magnetic field ( B Earth = 62.1 μT) is directed toward the north and is inclined downward at an angle θ to the horizontal. Find the magnitude of the magnetic force on a length L of the line due to Earth's field. Please also calculate a value numerically for I = 2000 A, = 73.0º and L = 100 m, since it’s useful to think about real world numbers. Since the line is running south to north, it has an angle θ relative to the Earth’s field. Thus: ( ) sin IF I L B = FL B G For the numbers given, ( ) ( )( )( )() sin 2000 A 100 m 62.1 sin 73º 12 N FI L B T θµ == = , which is a small force. For comparison, if the cable has a mass of 300 kg (about what a one cm radius copper wire of that length will weigh) it will feel a force of about 3000 N due to gravity. Of course most power lines are AC so this will also be an oscillating, not unidirectional force. Problem 3: Triangular Loop A current loop, carrying a current I , is in the shape of a right triangle with sides 3, 4, and 5 cm. The loop is in a uniform magnetic field B whose direction is parallel to the current in the 5 cm side of the loop. Find the magnitude of (a) the magnetic dipole moment of the loop in amperes-square meters and (b) the torque on the loop. The loop has an area of 2 1 2 base height 6 cm A =⋅ = so its dipole moment is (a) 2 6 c m I µ (b) The torque is 2 6 cm (in the plane of the loop, perpendicular to the hypotenuse) IB =×= τ μB G
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PS08 Solutions-2 Problem 4: A charge of mass m and charge q > 0 is at the origin at t = 0 and moving upward with velocity ˆ V = V j G . Its subsequent trajectory is shown in the sketch. The magnitude of the velocity V = V G is always the same, although the direction of V G changes in time. (a) For y > 0 , this positive charge is moving in a constant magnetic field which is either into the page or out of the page. Which is it? Into the Page (b) Derive an expression for the magnitude of the magnetic field for y > 0 . We just balance centripetal and magnetic force: 2 q V Bm VR Vq R =⇒ = (c) For y < 0 , the charge is moving in a different constant magnetic field. What is that magnetic field (magnitude and direction)?
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ps08sol - MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department...

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