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SR1ZFsNJI7da_1210016946_jwk572

# SR1ZFsNJI7da_1210016946_jwk572 - homework 17 KIM JI Due Apr...

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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- cury’s 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 = m v 2 r , so the field is B = m v q r = m q r radicalbigg 2 K m = 1 q r 2 m K . = 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 ˆ direction) and a B uniform mag- netic field (aligned with the ± ˆ 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: ˆ ı is in the direction + x (to the right), ˆ is in the direction + y (up the page), and ˆ 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

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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 = m v 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 0 . 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 , θ = 28 , A = 38 cm 2 = 0 . 0038 m 2 , and τ = 0 . 15 N m . The magnetic force on the current is vector F = I vector × vector B and the torque is vector τ = vector r × vector F , so the torque on the loop due to the magnetic field is τ = 2 F r cos θ = ( N I ℓ B ) w cos θ = N I B ( ℓ w ) cos θ = N I B A cos θ , where A is the area of the loop and θ is the angle between the plane of the loop and the magnetic field.
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SR1ZFsNJI7da_1210016946_jwk572 - homework 17 KIM JI Due Apr...

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