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In Fig 5.4 (b), the magnetic needle has magnetic moment 6.7 xx 10^(-2) Am^(2) and moment of inertia g=7.5 xx 10^(-6) Kg m^(-2). It performs 10 complete oscillation in 6.70s. What is the magnitude of the magnetic field? |
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Answer» SOLUTION :The time period of oscillation is, `T=(6.70)/(10)=0.67s` From Eq. (5.5) `B=(4pi^(2)g)/(m T)` `=(4 xx (3.14)^(2) xx 7.5 xx 10^(-6))/(6.7 xx 10^(-2) xx (0.67)^(2))` =0.01T |
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