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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. `1xx10^(-2)J//T`B. `2.08xx10^(-2)J//T`C. `3.08xx10^(-2)J//T`D. `1.52xx10^(-2)J//T` |
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Answer» Correct Answer - B Relation for dipole moment is, `M=IxxV`. Volume of the cylinder `V=pir^(2)l`, where r is the radius and `l` is the length of the cylinder. Then dipole moment `M=Ipir^(2)l=(5.30xx10^(3))xx22/7xx(0.5xx10^(-2))^(2)(5xx10^(-2))` `=2.08xx10^(-2)J//T` |
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