SR1ZFsNJI7da_1210016946_jwk572

SR1ZFsNJI7da_1210016946_jwk572 - homework 17 KIM, JI Due:...

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Unformatted text preview: homework 17 KIM, JI Due: Apr 7 2008, 4:00 am 1 Question 1, chap 30, sect 1. part 1 of 1 10 points A cosmic-ray proton in interstellar space has an energy of 1 . 4 MeV and executes a circu- lar orbit having a radius equal to that of Mer- curys orbit around the sun (5 . 8 10 10 m). The proton has a charge of 1 . 60218 10 19 C and a mass of 1 . 67262 10 27 kg. What is the magnetic field in that region of space? Correct answer: 2 . 94761 10 12 T (tolerance 1 %). Explanation: Let : q = 1 . 60218 10 19 C , m = 1 . 67262 10 27 kg , r = 5 . 8 10 10 m , and K = 1 . 4 MeV = 1 . 4 10 6 eV . From the kinetic energy K , we get the speed of the proton is v = radicalbigg 2 K m . The centripetal force is q v B = mv 2 r , so the field is B = mv q r = m q r radicalbigg 2 K m = 1 q r 2 mK . = 1 (1 . 60218 10 19 C) (5 . 8 10 10 m) radicalBig 2 (1 . 67262 10 27 kg) (1 . 4 10 6 eV) = 2 . 94761 10 12 T . Question 2, chap 30, sect 1. part 1 of 1 10 points A device (source) emits a bunch of charged ions (particles) with a range of ve- locities (see figure). Some of these ions pass through the left slit and enter Region I in which there is a vertical uniform electric field (in the direction) and a B uniform mag- netic field (aligned with the k-direction) as shown in the figure by the shaded area. q m Region of Magnetic Field B + V d x y z r Region I Region II Figure: is in the direction + x (to the right), is in the direction + y (up the page), and k is in the direction + z (out of the page). The ions that make it into Region II are observed to be deflected downward and then follow a circular path with a radius of r . The charge on each ion is q. What is the mass of the ions? 1. m = q E 2 r B 2. m = q B r E 3. m = q E 2 r B 4. m = q B r E 5. m = q E r B 6. m = q E r B 2 7. m = q B 2 r E 8. m = q B 2 r E correct 9. m = q B r E 2 homework 17 KIM, JI Due: Apr 7 2008, 4:00 am 2 10. m = q E r B Explanation: Since the electric and magnetic forces on the ion are equal, q E = q v B v = E B . The radius of a circular path taken by a charged particle in a magnetic field is given by r = mv q B . m = q B r v = q B r E B = q B 2 r E Question 3, chap 30, sect 3. part 1 of 1 10 points A small rectangular coil composed of 51 turns of wire has an area of 38 cm 2 and carries a current of 2 A. When the plane of the coil makes an angle of 28 with a uni- form magnetic field, the torque on the coil is . 15 N m. What is the magnitude of the magnetic field? Correct answer: 0 . 438301 T (tolerance 1 %). Explanation: Let : N = 51 turns , I = 2 A , = 28 , A = 38 cm 2 = 0 . 0038 m 2 , and = 0 . 15 N m ....
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This note was uploaded on 05/05/2008 for the course PHY 303L taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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SR1ZFsNJI7da_1210016946_jwk572 - homework 17 KIM, JI Due:...

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