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A longdielectriccylinderof radius Rus staticallyplartizedso thatat all the its pointsthe polarizationis equalto P = alpha x , wherealphais a positiveconstant, and ris the distancefrom the axis. Thecylinderis set into ratationabout its axiswith anangularvelocityomega. FInd themagnetic inductionBat the centreof the cylinder. |
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Answer» Solution :Because of polarization a space charge is present within the cylinder. It's density is `rho_(p) = -"div" vec(P) = -2 alpha` Since the cylinder as a whole is neutrala surface chargedensity `sigma_(p)` must be present on the surfaceof the cylinderalso. This has the magnitude(algebratically) `simga_(p) xx 2pi R = 2 alpha pi R^(2)` or, `sigma_(p) = alpha R` When the cylinder rotes, currents are set upwhich give riseto magenticfields. Thecontributesof `rho_(p)` and `sigma_(p)` can becalculated separaelyand then added. For the surface chargethe current is (for a particular element) `alpha R xx 2pi R dx xx (omega)/(2pi) = alpha R^(2) omega dx` Its contributionto the magentic fieldat the centre is `(mu_(0) R^(2) (alpha R^(2) omega dx))/(2(x^(2) + R^(2))^(3//2))` andthe total magentic field is `B_(s) = int_(oo)^(-oo) (mu_0) R^(2) (alpha R^(2) omega dx)/(2 (x^(2) + R^(2))^(3//2)) = (mu_(0) alpha R^(4) omega)/(2) int_(-oo)^(oo) (dx)/((x^(2) + R^(2))^(3//2)) = (mu_(0) alpha R^(4) alpha)/(2) xx (2)/(R^(2)) = mu_(0) alpha R^(2) SIGMA` As FPR the volume charge density considera circle of radius`r`, radial thickness `dr` and length`dx`. The currentis `-2A xx 2pi r dr dx xx (omega)/(2pi) = -2 alpha r dr omega dx` The total magnetic field due on the the volumecharge distribution is `B_(v) = -int_(0)^(R) dr int_(-oo)^(oo) dx 2pi r omega(mu_(0) r^(2))/(2(x^(2) + r^(2))^(3//2)) = -int_(0)^(R) alphamu_(0) omegar^(3) dr int_(-oo)^(oo)dx (x^(2) + r^(2))^(3//2)` `= -int_(0)^(R) alpha omega r dr xx 2 = -mu_(0) alpha omegaR^(2)` so,`B = B_(s) + B_(v) = 0` |
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