PHY132 Sp04 Midterm 2 Solutions

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

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon