HW 07-solutions-2

# Because the currents are in op posite directions the

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straight wire.); because the currents are in op- posite directions, the forces on them will be in opposite directions. Their contributions to the net force cancel. The net force on the loop is then F loop = F AB - F CD = I 2 μ 0 I 1 2 π a - I 2 μ 0 I 1 2 π ( a + b ) = μ 0 I 1 I 2 2 π b a ( a + b ) . A B C D I 1 I 2 I 2 I 2 I 2 I 2 × B 1 F AB I 2 × B 1 F CD Here, a positive force is in the up direction, and a negative force is in the down direction. So, the direction of the net force is up. 002 10.0 points A static uniform magnetic field is directed out of the page. A charged particle moves in the plane of the page following a clockwise spiral of increasing radius as shown. B B Neglect the e ff ect due to gravity. What is a reasonable explanation? 1. The charge is negative and speeding up. 2. The charge is neutral and speeding up. 3. The charge is neutral and with a constant speed. 4. The charge is positive and slowing down. 5. The charge is neutral and slowing down. 6. The charge is positive and speeding up. correct 7. None of these 8. The charge is negative and with a con- stant speed. 9. The charge is positive and with a constant speed. 10. The charge is negative and slowing down. Explanation: We know that when a charged particle moves in a uniform magnetic field with a constant speed, it undergoes a circular mo- tion with the centripetal force provided by the magnetic force, namely m v 2 r = q v B , so we know that the radius is in fact propor- tional to the speed, r = m q B v . Since the particle follows a spiral of increasing radius, we can judge that it is speeding up. The magnetic force F = q v × B must be in the direction for the centripetal force - ˆ r (pointed inward) of this particle in clockwise circular motion. Since v × B is in the negative ˆ r direction, the particle has a positive charge. 003 (part 1 of 2) 10.0 points

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karna (pk4534) – HW 07 – li – (59050) 3 A circular current loop of radius R is placed in a horizontal plane and maintains a current I . There is a constant magnetic field B in the xy -plane, with the angle α (0 < α < 90 ) defined with respect to y -axis. The current in the loop flows clockwise as seen from above. In this problem we determine the torque vector τ which the field exerts on the current loop. I z ˆ k x ˆ ı y , ˆ B α What is the direction of the torque vector τ ? 1. τ = ˆ ı + ˆ k 2 2. τ = - ˆ 3. τ = ˆ ı + ˆ sin α 4. τ = ˆ k - ˆ ı 2 5. τ = - ˆ k 6. τ = ˆ k - ˆ sin α 7. τ = +ˆ ı 8. τ = +ˆ 9. τ = + ˆ k correct 10. τ = - ˆ ı Explanation: Basic Concepts: Torque on a current loop due to a magnetic field. μ B I z ˆ k x ˆ ı y , ˆ α Solution: We know that torque is τ = μ × B = μ ( - ˆ ) × [ B x (+ˆ ı ) + B y ( - ˆ )] and that ˆ × ˆ = 0 and - ˆ × ˆ ı = ˆ k , so τ = μ B x ˆ k , and the direction of the torque is + ˆ k . This agrees with the answer from the right-hand rule.

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