006 part 1 of 3 100 points refer to a long straight

This preview shows 3 out of 5 pages.

006 (part 1 of 3) 10.0 points Refer to a long, straight wire carrying con- stant current I . What can be concluded about the magni- tude of the magnetic field at distance a from the wire? ( α is read “is proportional to”) 1. B I a 2 2. B I a 3. B I 4. B a 5. B I a correct Explanation: Ampere’s Law tells us that therefore B · 2 π a = μ 0 I B = μ 0 I 2 π a 1 a . 007 (part 2 of 3) 10.0 points If a stationary charge Q were located a dis- tance a from the wire, the force on that charge due to the magnetic field of the wire would be 1. infinite. 2. 0. correct 3. dependent on current I. 4. dependent on distance a . 5. in the direction of the wire. Explanation: Magnetic force on a charge is proportional to the velocity of the charge ( F B = q v B ), so the force is zero when the velocity is zero. 008 (part 3 of 3) 10.0 points If an identical wire were parallel to the first wire with a distance a between them and the second wire also carried the same current I (in the same direction), then the force on the second wire would be 1. 0. 2. dependent on the value of a only. 3. repulsive from the first wire. 4. attractive to the first wire. correct
Image of page 3

Subscribe to view the full document.

maini (nm7637) – hw10 – Shneidman – (12108) 4 5. time-varying. Explanation: The magnetic force on a segment of wire of length L carrying current I is given by F B = L vector I × vector B , where vector B in this case is the field due to the other wire and vector I points in the direction of the current. This field B 1 due to the first wire points perpendicular to the second wire (where currents I 1 and I 2 point into the page). I 1 I 2 F 12 B 1 The right hand rule for cross-products de- termine that the force F 12 on the second wire due to the first will point toward the first wire. 009 10.0 points Consider a long wire and a rectangular current loop. A B C D I 1 b a I 2 Determine the magnitude and direction of the net magnetic force exerted on the rectan- gular current loop due to the current I 1 in the long straight wire above the loop. 1. vector F = μ 0 I 1 I 2 2 π ( a - b ), right 2. vector F = μ 0 I 1 I 2 2 π parenleftbigg a a + b parenrightbigg , down 3. vector F = μ 0 I 1 I 2 2 π ( a - b ),left 4. vector F = μ 0 I 1 I 2 2 π bracketleftbigg b a ( a + b ) bracketrightbigg , up correct 5. vector F = μ 0 I 1 I 2 2 π a b , down 6. vector F = μ 0 I 1 I 2 a 2 π ℓ ( b - a ), up 7. vector F = μ 0 I 1 I 2 2 π ( a + b ), up 8. vector F = μ 0 I 1 I 2 2 π parenleftbigg a b a + b parenrightbigg , down Explanation: To compute the net force on the loop, we need to consider the forces on segments AB , BC , CD , and DA . The net force on the loop is the vector sum of the forces on the pieces of the loop. The magnetic force on AB due to the straight wire can be calculated by using vector F AB = I 2 integraldisplay B A dvectors × vector B . In order to use this, we need to know the magnitude and direction of the magnetic field at each point on the wire loop. We can apply the Biot-Savart Law. The result of this is that the magnitude of the magnetic field due to the straight wire is B = μ 0 I 1 2 π r , and the direction of the magnetic field is given by the right hand rule; the field curls around the straight wire with the field coming out of the page above the wire and the field going into the page below the wire. We can now find the force on the segment AB ; applying
Image of page 4
Image of page 5
You've reached the end of this preview.
  • Spring '08
  • moro
  • Current, Magnetic Field, Wire

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern