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Problem Set 8 Solutions

Problem Set 8 Solutions - MASSACHUSETTS INSTITUTE OF...

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MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics 8.02 Fall 2007 Problem Set 8 Solutions Problem 1: Charges in a Uniform Field An electron with velocity ˆ ˆ x y v v + i j moves through a uniform magnetic field ˆ ˆ x y 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 ˆ ˆ 0 0 x y x y y x x y q q v v q v B v B B B = × = = i j k F v B G G G k So the magnitude of the force is ( ) x y y x F e 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 = 3000 A, θ = 78.0º 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 I F ILB θ = × = F L B G G G For the numbers given, ( ) ( )( )( ) ( ) sin 3000 A 100 m 62.1 sin 78º 18 N F ILB T θ μ = = = , 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 30, 40, and 50 cm. The loop is in a uniform magnetic field B whose direction is parallel to the current in the 50 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 600 cm A = = so its dipole moment is (a) 2 1200 cm I μ = (b) The torque is 2 600 cm (in the plane of the loop, perpendicular to the hypotenuse) IB = × = τ μ B G G G PS08 Solutions-1
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