859_Physics ProblemsTechnical Physics

# 859_Physics ProblemsTechnical Physics - Chapter 30 a INR 2...

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Chapter 30 199 (a) B INR u Ru R IN a aR a a a = + = + −+ L N M M M O Q P P P µµ 0 2 22 2 0 22 2 2 AA A A A a f (b) If A is much larger than R and a = 0, we have B IN IN ≅− L N M M O Q P P = 0 2 0 2 0 2 A A A A . This is just half the magnitude of the field deep within the solenoid. We would get the same result by substituting a = A to describe the other end. P30.33 The field produced by the solenoid in its interior is given by Bi i =− = × F H G I K J µπ 0 7 2 41 0 30 0 10 15 0 nI ± . . ± eje j af ej TmA m A × 565 10 2 . ± T The force exerted on side AB of the square current loop is FL B j i B AB I bg e j e j =×= × × × −− 0 200 2 00 10 5 65 10 .. ± . ± A m T Fk B AB 226 10 4 . ± N Similarly, each side of the square loop experiences a force, lying in the plane of the loop, of 226 N directed away from the center µ . From the above result, it is seen that the net torque exerted on the square loop by the field of the solenoid should be zero. More formally, the magnetic dipole moment of the square loop is given by
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## This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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