PHY132 Sp04 Midterm 2 Solutions

# University Physics with Modern Physics with Mastering Physics (11th Edition)

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PHY 132 Midterm 2 Mar 31, 2004 Version A Put FULL NAME, ID#, and EXAM VERSION on the front cover of the BLUE BOOKLET! To avoid problems in grading: do all problems in order , write legibly , and show all work for partial credit ! Note: make sketches where appropriate and show your axis conventions. Do not forget units, number of signifi- cant digits, and check your results for consistency. This exam will last 1.5 hr. Success! 1. A copper rod of length D= 0.50 m is pulled with constant velocity v = 4.0 m/s, perpendicular to a uniform and constant magnetic field B = 0.25 T, see the Figure. The rod glides without friction on a horizontal U- shaped conductor as shown. Give directions in terms of up , down , left , right , CW , CCW , into and out of the paper. Show all work ! (60 points) a. Calculate the force (magnitude and direction) on an elec- tron (charge –e) in the rod. Solution: (10 points) F = q v × B = – e vB (down) = 1.6×10 –19 N (up) ( B : 1.6×10 –19 N (down)) a v b D c d B b. Calculate the magnitude and direction of the induced emf. Solution: (10 points) | E | = DvB = 0.50 V; direction: CW ( B : = LvB = 0.40 V; CCW) c. Assume the resistance of the circuit abcd is 0.10 , constant and dominated by the contact resistance at points a and b . Calculate the magnitude of the induced current I induced in the circuit and give its direction. Solution: (10 points) I = | E |/R = 5.0 A; direction: CW ( B : 2.0 A, CCW) d. Calculate the magnetic force (magnitude and direction) on the rod due to the induced current. Solution: (10 points) F M = I D × B = IDB (right) = 0.625 N (right) ( B : = ILB= 0.160 N (right)) e. Calculate the power dissipated in the circuit loop. Solution: (10 points) P = I 2 R = 2.5 W ( B : 0.80 W) f. Calculate the magnitude of the external force pulling on the rod. Solution: (10 points) Action is reaction: F pull = – F M = 0.625 (left) ( B : 0.160 (left)) 2. Consider two long conducting coaxial cylinders . The inner cylinder is solid and has radius a= 3.0 cm. It carries a uniform current I = 900 A in the direction shown. The outer conducting cylinder has inner radius b= 10 cm, and outer radius c= 11 cm, and carries an equal current in the opposite di- rection. The space between the cylinders is filled with insulating plastic ( K m = 1.0). Give directions in terms of up , down , left , right , into and out of the the paper. Show all work! (30 points) Q a b c I P a. Use Ampere’s Law to calculate the magnetic field for radial distance i ) r<a , and for ii ) a<r<b . Note: for full credit you must start from Ampere’s Law! Solution: (14 points) i ): 2 0 0 0 2 2 2 ( ) ( ) 0.200 T/m 2 encl r a I I d r B r I r B r r r a a µ π µ µ π π π < = = = = = B l v

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