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A cylindrical rod magnet has a length of 5 cm and a diameter of 1 cm. It has a unifirm magnetisation of `5.30xx10^(3) Amp//m^(3)`. What its magnetic dipole moment?A. `20.8mJT^(-1)`B. `10.8mJT^(-1)`C. `5.8mJT^(1)`D. `30.8mJT^(-1)` |
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Answer» Correct Answer - A `M(m)/(V)` or `m=MV` `m=Mxx(pi r^(2)l)` `=(5.3xx10^(3))(3.14)(5xx10^(-3))^(2)(5xx10^(-2))J//T =20.8xx10^(-3)J//T=20.8mJ//T` |
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