Physics 112 Homework 6 (solutions) (2004 Fall) SolutionstoHomeworkQuestions 6 Chapt20, Problem-1: A magnetic field of strength 0.30 T is directed perpendicular to a plane circular loop of wire of radius 25 cm. Find the magnetic flux through the area enclosed by this loop. Solution: The magnetic flux through the area enclosed by the loop is ! B = BAcos " = B #r 2 ( ) # ( 0.25 m ) 2 ( )cos 0°= 0.30 T [ ]= 5.9 ! 10"2 T #m 2 Chapt20, Problem-3: Consider a place where Earth’s magnetic field has strength of 0.520x10 –4 T and makes an angle of 62.0° with the horizontal. At this location a magnetic compass points toward true north. What is the magnetic flux in a rectangular loop of wire that is 15.0 by 25.0 cm and is lying on a table What is the magnetic flux in this loop if it is mounted vertically on a north wall On an east wall Solution: When the loop lies on a horizontal table, the angle between the field and the normal to the plane of the loop is ! = 90.0° " 62.0°=28.0° . The flux through the area enclosed by the loop is then ! B = BAcos " = 0.520 !10 "4 T ( )( 0.25 m) ( ) 0.150 m [ ]cos 28.0°= 1.72! 10"6 T #m 2 When the loop is mounted vertically on a north wall, ! = 62.0° and ! B = ( 0.520 "10 #4 T) [( 0.150 m )( 0.25 m) ]cos 62.0°= 9.15! 10"7 T #m 2 When the loop is mounted vertically on an east wall, ! = 90.0° , so ! B = BAcos " = BAcos 90.0°= 0 Chapt20, Problem-11: A strong electromagnet produces a uniform field of 1.60 T over a cross-sectional area of 0.200 m 2 . We place a coil having 200 turns and a total resistance of 20.0 ! around the electromagnet. We then smoothly decrease the current in the electromagnet until it reaches zero in 20.0 ms. What is the current induced in the coil Solution: The magnitude of the induced emf is 200( 1.60 T % 0) 0.200 m 2 ! = "# B "t = N ( "B )Acos $ = "t Thus, the current induced in the coil is ( )cos 0° 20.0 &10 -3 s = 3.20& 10 3 V I = ! R = 3.20 "103 V 20.0 # = 160 A 1

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