A 50 gram meterstick is placed on a fulcrum at its 50 cm mark. A 20 gram mass is attached at the 12 cm mark. Where should a 40 gram mass be attached so that the meterstick will be balanced in rotational equilibrium

The 40g mass will be attached at 69 cm

Explanation:

First, make a sketch of the meterstick with the masses placed on it;

               ↓                    Δ                      ↓

             20 g50 cm40g

                         38 cm                  y cm  

Apply principle of moment;

sum of clockwise moment = sum of anticlockwise moment

40y = 20 (38)

40y = 760

y = 760 / 40

y = 19 cm

Therefore, the 40g mass will be attached at 50cm + 19cm = 69 cm

              12cm             50 cm              69cm

               ↓                    Δ                      ↓

             20 g50 cm40g

                         38 cm                 19 cm                                              

the higher the current the higher the mag field around it the higher the influence on the compass

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach