_016 - Solution for Homework 17 Faradays Law Solution to...

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Unformatted text preview: Solution for Homework 17 Faradays Law Solution to Homework Problem 17.1(Motional EMF Across Bar) Problem: A metal bar of length = 3cm is dragged through our . 2T lab magnets at a velocity of v = 5 m s in the direction shown in the figure at the right. Calculate the emf induced across the bar. Select One of the Following: (a) 0V (b) 1V (c) . 15V (d-Answer) . 03V (e) . 004V Magnetic Field Into Page v x y z Solution The motional emf of a wire moving through a magnetic field is given by emf = integraldisplay C vectorv vector B d vector Write vector B = B z and vectorv = v y . The cross product vectorv vector B = ( v y ) ( B z ) = v B y z = v B x The path element along the bar is d vector = xdx and therefore the dot product is vectorv vector B d vector = v B x ( xdx ) = v B dx The integral is then emf = v B integraldisplay dx = v B = (5 m s )(0 . 2T)(0 . 03m) = 0 . 03V where is the length of the bar. Total Points for Problem: 4 Points Solution to Homework Problem 17.2(Comparing Potential Difference Across Solenoids With Different Numbers of Wrappings) Problem: When a magnet is moved in and out of a solenoid with 3 wrappings of wire, a potential difference V is produced across the solenoid. What is the voltage produced in terms of V when the solenoid has 6 wrappings? Select One of the Following: (a-Answer) The voltage across the solenoid increases to 2 V . 1 (b) The voltage across the solenoid increases to 4 V . (c) The voltage across the solenoid decreases to V/ 2 . Solution The turns of wire each have the same voltage, and adding an additional turn of wire adds that same amount of voltage to the solenoid. Thus, doubling the wrappings doubles the potential difference across the solenoid. Total Points for Problem: 3 Points Solution to Homework Problem 17.3(Motional EMF Across Bar ) Problem: A metal bar of length = 3cm is dragged through our . 2T lab magnets at a velocity of v = 5 m s in the direction shown to the right. Calculate the emf induced across the long dimension of the bar. Select One of the Following: (a-Answer) 0V (b) 1V (c) . 15V (d) . 03V (e) . 004V Magnetic Field Into Page v x y z Solution The motional emf of a wire moving through a magnetic field is given by emf = integraldisplay C vectorv vector B d vector Write vector B = B z and vectorv = v x . The cross product vectorv vector B = ( v x ) ( B z ) = v B x z =- v B y The path element along the bar is d vector = xdx and therefore the dot product is vectorv vector B d vector =- v B y ( xdx ) = 0 The integral is then emf = 0 . Total Points for Problem: 4 Points Solution to Homework Problem 17.4(EMF of Growing Square) Problem: A square loop of wire with N turns is growing in a constant uniform magnetic field, B , as shown below. The length of one of the sides of the square, ( t ) , is increasing with time as the function ( t ) = t 2 .....
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_016 - Solution for Homework 17 Faradays Law Solution to...

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