Question

You are in the lab and are given two rods set up so that the top rod is directly above the bottom one. The two straight rods 50-cm long are separated by a distance of 1.5-mm each with 15 A of current passing through them. Assume that the bottom rod is fixed and that the top rod is somehow free to move vertically but not horizontally. How much must the top rod weigh so that the system is at equilibrium

Answer 1

Step-by-step explanation:

The magnetic force due to lower rod must be equal to weight of upper rod for equilibrium .

magnetic field due to lower rod on upper rod

= ( μ₀ / 4π ) x(2i / r ) , i is current , r is distance between rod

= 10⁻⁷ x 2 x 15 / 1.5 x 10⁻³

= 20 x 10⁻⁴ T

force on the upper rod

= B i L , B is magnetic field , i is current in second rod and L is its length

= 20 x 10⁻⁴ x 15 x .50

= 150 x 10⁻⁴ N

= .015 N

This force can balance a wire having weight equal to .015 N .

= .00153 kg

= 1.53 g .

wire should weigh 1.53 g .