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A magnetic needle placed in uniform magnetic field has magnetic moment 6.7 xx 10^(-2) "Am"^(2), and moment of inertia of 15 xx 10^(-6) k m^(2). It performs 10 complete oscillations in 6.70 s. What is the magnitude of the magnetic field ? |
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Answer» Solution :Dipole moment ` = 6.7 xx 10^(-2) "Am"^(2)` Moment of inertia `I= 15 xx 10^(-6) "KGM"^2` Periodic time T `= ("time")/("No. of oscillations")` `- (6.7)/(10) -0.67` s The period of magnetic needle placed in uniform magnetic field, `T= 2pi sqrt((I)/( MB))` `therefore T^(2) = (4 pi^(2) I)/( mB)` `therefore B= (4PI^(2) I)/( mT^(2) )` `= (4 xx (3.14)^(2) xx 15 xx 10^(-6) )/( 6.7 xx 10^(-2) xx (0.67)^(2) )` `= 196.69 xx 10^(-4)` `therefore B ~~ 0.02` T |
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