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| 1. |
A magnet vibrates in a magnetic field of strength 10^(-4) T. If the moment of the magnet is 10^(-1)A-m^2 and the moment of inertia about the axis of rotation is 10^(-5) kg m^(2), then find the time period of oscillation in seconds? |
| Answer» Solution :Period of OSCILLATION , `T= 2pi SQRT((I)/(MB))= 2pi sqrt((10^(-5))/(10^(-1) xx 10^(-4)))= 2pi `SEC. | |