Unformatted text preview: PHY2054 Spring 2008 l2. (Test Bank 30) A horizontal current—carrying wire of length 50 cm and mass 0.2 kg is placed
perpendicular to a uniform horizontal magnetic ﬁeld of magnitude 027 T. For what current will the wire ﬂoat due to the magnetic force acting upward on it? (g = 9.8 m/sz) v V/
, \ “ ' ._ D
@45 A K, .j _ " ‘
(2) 1.5A K 4, ,._(Q\
(3)3.9A \ // ~
(4) 25 mA ‘9
(5) 150 mA V Balance of gravitational and magnetic forces leads to the equation mg = BiL. Solving for 1'
yields 14.5 A. 13. (i eat Bank 39) A rectangular coil (0.20 m x 0.80 at) has 200 turns and is placed in a uniform
magnetic ﬁeld of 0.30 T. If the orientation of the coil is varied through all possible positions, the
maximum torque on the coil by magnetic forces is 0.080 Nm. What is the current in the coil? @3 mA
.7 A (3) 2.5 mA
(4) 1.0 A
(5) 50 mA The maximum torque on the coil is r = NAiB, where N = 200, A = 0.3 x04 = 0.12 m2, 1 =
0.080 and B = 0.30 T. Solving for i yields 8.3 m. M W“ l4}. (Cimp. 19.31) A singly charg  sAMﬂu  = . mass of 2.68 x 10'26 kg. Alter being accelerated through a potential diﬂ‘erence of 232 V, the ion enters a magnetic ﬁeld of 0.590 T, in
a direction mndicular to the ﬁeld. Calculate the radius of the path of the ion in the ﬁeld. (1) 1.49 cm ' , * (2) 1.77 cm Fm 0k Owl
(3) 8.82 mm 0W U ‘ V‘ ‘ ’
(4) 4.5 cm (5) none of these The radius is found by equating the centripetal force to the magnetic force, (or mv2 /r = evB ,
which yields r = mv/ eB. From the acceleration over a potential, we can also write émvz = eV. Putting these together yields the equation r = JZmV/ e / B . Solving for r yields 1.49 cm. V cm}.
(L1?) 5 ‘5 VmﬂAi/ ...
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 Spring '09
 DUNNAM

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