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Magnetic torque on sphere on inclined plane

  1. Feb 24, 2008 #1
    [SOLVED] Magnetic torque on sphere on inclined plane

    1. The problem statement, all variables and given/known data
    a nonconducting sphere has mass 80 g and radius 20 cm. a flat, compact coil of wire with 5 turns is wrapped tightly around it, with each turn concentric with the sphere. the sphere is placed on an inclined plane that slopes downward to the left, making an angle theta with the horizontal so that the coil is parallel to the inclined plane. a uniform magnetic field of .35 Tesla vertically upward exists in the region of the sphere. what current in the coil will enable the sphere to rest in equilibrium on the inclined plane? show that the result does not depend on the value of theta.

    2. Relevant equations F = q(v x B)(nAL), Tau = mu x B, and mu of coil = NIA

    3. The attempt at a solution
    mu of coil = NIA = 5 (I)(2piR) = 5(2pi)(20)(I)
    so mu of coil = 100 (pi)(I)

    then Tau = mg sin theta = mu x B sin theta = NIA (B) sin theta
    so I = mg sin theta/(NAB sin theta)
    substituting I = .08(6.673 x 10 -11)/(5 pi)(2 x 10 -2) squared (.35)
    = (.53384)(x 10 -11)/(21.99 x 10 -4) = 243 x 10 -11

    but the book answer is .713 Amps CCW, so i am off by a factor of a trillion or so again :-(
  2. jcsd
  3. Feb 26, 2008 #2
    i needed the formula Torque = r cross mg, and i needed to use little g acceleration and not big G gravitational constant. that solves it.
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