Practicing Success
The time period of a small magnet in a horizontal plane is T. Another magnet B oscillates at the same place in a similar manner. The size of two magnets is the same but the magnetic moment of B is four times that of A. The time period of B will be- |
$\frac{T}{4}$ $\frac{T}{2}$ 2T 4T |
$\frac{T}{2}$ |
$ \frac{T_A}{T_B} = \sqrt{\frac{M_B}{M_A}} $ $ T_B = T\sqrt {\frac{M_A}{4M_A}} = \frac{T}{2}$ |