PHYSICS MAGNETICS PLEASE ANSWER

Question

a metal rod of mass 10 gm and length 25 cm is suspended on two springs one at each end of the rod .the springs are extended by 4 cm in this situation.when a current 20A passes through the rod it rises by 1 cm. determine the magnetic field assuring acceleration due to gravity to be 10m/s2

Answer 1

The spring constant of the springs (combined) is m*g/s = 0.01 kg * 10 m/s² / 0.04 m = 2.5 N/m.  To rise by 0.01m the force is 0.01*2.5 = 0.025 N

The force from a current is I*L x B, if the magnetic field is at right angles to the current, then

B = F/(I*L) = 0.025/(20*0.25)  = 0.005 T  

Answer 2

Can you simlify your question

Answer 3

The spring constant of the springs (combined) is m*g/s = 0.01 kg * 10 m/s² / 0.04 m = 2.5 N/m. To rise by 0.01m the force is 0.01*2.5 = 0.025 N

The force from a current is I*L x B, if the magnetic field is at right angles to the current, then

B = F/(I*L) = 0.025/(20*0.25) = 0.005 T  

Answer 4

is this some iit question??? man.. i think with corresponding formulas u can do it.. see some phy texts