Two straight rods 60 cm long and 2.0 mm apart in a current balance carry currents of 18 A each in opposite directions. What mass must be placed on the upper rod to balance the magnetic force of repulsion?

M = 1.98g

Explanation:

Please see the attachment below.

For the mass to balance the force of repulsion, the gravitational force on the mass (its weight) must equal the magnetic force of repulsion between the straight rods.

So

Mg = μo ×I ×I' × L/2πr

M = μo ×I ×I' × L/2πrg

Given

I = I' = 18A

L = 60cm = 0.6m

r = 2mm = 2×10-³m

μo = 4π×10-⁷Tm/A

g = 9.8m/s²

M = (4π×10-⁷ × 18² × 0.6)/(2π×2×10-³ ×9.8) = 1.98 ×10-³ kg.

r = ρ x length / area  

ρ = r x area / length  

ρ is in ohm-meters  

in basic units, it is m³kg/s³a²  

i think this is it.