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A magnetic needle of magnetic moment 6.7×10−2 Am2 and moment of inertia 7.5×10−6 kg m2 is performing simple harmonic oscillations in a magnetic field of 0.01 T.  Time taken for 10 complete oscillations is ?

Answers (1)

s we learnt in

Time period of Oscillating Bar Magnet -

T=2\pi \sqrt{\frac{I}{MB_{H}}}

- wherein

I = Moment of Inertia of short bar magnet 

I =\frac{mL^2}{12}

M - Mass of bar magnet

 

 T=2\pi \sqrt{\frac{I}{MB_{H}}}    

I=7.5 \times 10^{-6} Am^{2}\\ M=6.7\times 10^{-2} kgm^{2}\\ B=0.1 T

Hence T = 6.65 s

Posted by

Safeer PP

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