EEL3472HA4S - HA4 r"/( 20 t0 Problcrn 5.2...

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Unformatted text preview: HA4 r"/( 20 t0 Problcrn 5.2 When a particle with charge q and mass rn is introduced into a medium with a uniform field B such that the initial velocity of the particle u is perpendicular to B(Fig. P5.2), the magnetic force exerted on the particle causes it to move in a circle of radius o. By equating F. to the centripetal force on the particle, determine a in terms of q, m, u, andB. @ @ @ @e fiigure P5.2: Particle of charge q projected with velocitv u into a medium with a uniform field B perpendicular to u (Problem 5.2). Solution: The centripetal force acting on the particle is given by 4: mu2fa. Equating ,I[ to F' given by Eq. (5.4), we have mu2 f a - quB sinl. Since the magnetic field is perpendicular to the particle velocity, sin 0 : l. Hence, a : mu /qB. @ @ @ @ @ @ Problem 5.tl Use the approach outlined in Example 5-2 to develop an expression for the magnetic field H at an arbitrary point P due to the linear conductor defined by the geometry shown in Fig. P5.8. If the conductor extends between zt : 3 m and z2 :'l m and carries a current I : 15 A, find H at P : (2, Q,0). PzQz) I P{') _ - - - : i p : ( r , Q , z ) l'igure P5.8: Current-carrying linear conductor of Problem 5.8. Sofutinn: The solution follows Example 5-2 up through Eq. (5.27), but the expressions for the cosines ofthe angles should be generalized to read as c o s 0 1 : , c o s 0 2 : instead of the expressions in Eq. (5.28), which are specialized to a wire centered at the origin. Plugging these expressions back into Eq. (5.27), the magnetic field is given as z - 2 2--_ { r'* (t - rz)" ":6*( For the specific geometry of Fig. , . 1 5 l - o - 3 0 - 7 I . - ^ . . ^ - 2 . ^ , H:q#1ffi- ffil:ottlx r0-3 (A/m) :$tt'+ (mA/m)' * + ( z - t r ) ' Prgblerl 5.ll) An infinitely long, thin conducting sheet defined over the space 0 S x < w and -*...
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EEL3472HA4S - HA4 r"/( 20 t0 Problcrn 5.2...

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