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# img062 - PHY2054 Spring 2008 l2(Test Bank 30 A horizontal...

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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 N-m. 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}. (Ci-mp. 19.31) A singly charg - sAM-ﬂu - = . 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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