UnitExam4Bsolutions

# UnitExam4Bsolutions - ID B UCSD Physics 2B Answer Section...

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ID: B 1 UCSD Physics 2B Unit Exam 4B Magnetism & Induction Answer Section MULTIPLE CHOICE 1. ANS: D r = mv qB = 1.67 × 10 27 kg Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3.7 × 3 m / s Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 1.60 × 19 C Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 3.22 × 3 T Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ = 12 mm TOP: CYCLOTRON RADIUS 2. ANS: A Applying Ampere's Law in a loop of radius 10 mm around the wire, B = μ 0 I 2 π r = 1.26 × 6 H / m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ A () 6.28 × 3 m Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ = 2.01 × 4 T = 201 × 6 T TOP: WIRE FIELD 3. ANS: D The magnitude is F ||= I l × B || = I lB sin ϑ = 1.5 A 3.0 m 1.0 T sin 28 = 2.1 N TOP: FORCE | WIRE 4. ANS: C F = q v × B whose magnitude F = qvB sin . Lay a coordinate system with east = i, north = j and up = k . Curl right-hand fingers from vector v to B . If the thumb points out of the page ( k ), then the angle = 180° + 35° = 215°. If the thumb points into the page (- k ), then the angle = 180° - 35° = 145°. Now solve for magnitude: F = =− 1.6 × C Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 2.5 × 5 m / s Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ 0.45 T 1.80 × 14 N For the projection & direction, sin 215 k 5.74 × 1 k while the other choice of orientation gives the same result: sin 145 k = 5.74 × 1 k 5.74 × 1 k The final answer is thus F 5.74 × 1 Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ × N Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ k = 1.03 × N k so the direction is "up" or out of the page . TOP: FORCE

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ID: B 2 5. ANS: D For the trajectory to be undisturbed, the combined net force must be zero: F = qE + vB () F = 0 E =− vB = 7.33 × 10 6 m s Ê Ë Á Á Á Á Á Á ˆ ¯ ˜ ˜ ˜ ˜ ˜ ˜ 8.31 × 5 T Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜ = 609 V m TOP: CROSSED FIELDS 6. ANS: C The magnetic moment of the coil μ = NIA = NI π R 2 Ê Ë Á Á Á ˆ ¯ ˜ ˜ ˜
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UnitExam4Bsolutions - ID B UCSD Physics 2B Answer Section...

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