InterviewSolution
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InterviewSolution
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Statement A : A proton has spin and magnetic moment just like an electron. But its effect is neglected in magnetism of material. Statement B : The order of magnitude of difference between the diamagnetic suceptibility of N_(2)(~5xx10^(-9))("STP")andCu(~10^(-5))" is "1.6xx10^(-4) Statement C : Suppose we went to verify the analogy between electrostatic and magnetostatic by an explicit experiment. Consider the motion of (i) electric dipole P in an electrostatic field E and (ii) magnetic dipole M in a magnetic field B. Set of conditions on E,B,p,M so that the two motions are verified to be identical. (Assume identical initial contiditions are (i) P=M/C (ii) PE = MB |
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Answer» A correct B correct C correct `M=(eh)/(4pim)orMprop1/m` `(because(eh)/(4pi)="constant")` `thereforeM_(p)/M_(e)=m_(e)/m_(p)=M_(e)/(1837M_(e))` `(becauseM_(p)=1837M_(e))` `rArrM_(p)/M_(e)=1/1837ltlt1` `rArrM_(p)ltltM_(e)` Thus, EFFECT of magnetic moment of proton is neglected as compared to that of electron. B) We know that Density of nitrogen `rho_(N_(2))=(28g)/(22.4L)=(28g)/(22400c c)` Also, density of copper `rho_(Cu)=(8g)/(22.4L)=(8g)/(22400c c)` Now, comparing both densities `rho_(N_(2))/rho_(Cu)=(28)/(22400)xx(1)/(8)=1.6xx10^(-4)` Also given `x_(N_(2))/X_(cu)=(5XX10^(-9))/(10^(-5))=5xx10^(-4)` We know that, `x=("Magnetisation(M)")/("Magnetic int ensity(H)")` = `M/(HV)=M/(H("mass"//"density"))=M_(p)/(Hm)` `thereforexproprho" ""Hence"," "x_(N_(2))/x_(Cu)=rho_(N_(2))/rho_(Cu)=1.6xx10^(-4)` Thus, the order of magnitude difference or major difference between the diamagnetic susceptibility of `N_(2)` and Cu is accounted for by the ratio of densities. C) Now, SUPPOSE that the angle between M and B is `theta` Torque on magnetic dipole moment M in magnetic field B. `tau.=MBsintheta` Two motions will be identical, if `pEsintheta=MBsintheta` `rArrpE=MB` But, `E=cB` `therefore` Putting this value in Eq. (i), `pcB=MB` `rArrp=M/c` |
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