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PhyIIHW7Solutions - hopkins(tlh982 HW07 criss(4908 This...

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hopkins (tlh982) – HW07 – criss – (4908) 1 This print-out should have 6 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A long cylindrical shell has a uniform current density. The total current flowing through the shell is 8 mA. The permeability of free space is 1 . 25664 × 10 - 6 T · m / A . 10 km 7 cm 2 cm The current is 8 mA . Find the magnitude of the magnetic field at a point r 1 = 4 . 5 cm from the cylindrical axis. Correct answer: 12 . 8395 nT. Explanation: Let : L = 10 km , r a = 2 cm = 0 . 02 m , r b = 7 cm = 0 . 07 m , r 1 = 4 . 5 cm = 0 . 045 m , I = 8 mA , and μ b = 1 . 25664 × 10 - 6 T · m / A . L r b r a The current I = 8 mA . Since the cylindrical shell is infinitely long, and has cylindrical symmetry, Ampere’s Law gives the easiest solution. Consider a circle of radius r 1 centered around the center of the shell. To use Ampere’s law we need the amount of current that cuts through this circle of radius r 1 . To get this, we first need to compute the current density, for the current flowing through the shell. J = I A = I π r 2 b π r 2 a = (8 mA) π [(0 . 07 m) 2 (0 . 02 m) 2 ] = 0 . 565884 A / m 2 . The current enclosed within the circle is I enc = π [ r 2 1 r 2 a ] · J = π [(0 . 045 m) 2 (0 . 02 m) 2 ] × (0 . 565884 A / m 2 ) = 0 . 00288889 A . Ampere’s Law, contintegraldisplay vector B · dvectors = μ 0 I enc B 2 π r 1 = μ 0 I enc B = μ 0 I enc 2 π r 1 = 1 . 25664 × 10 - 6 T · m / A 2 π (0 . 045 m) × (0 . 00288889 A) = 12 . 8395 nT .
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