Question

A horizontal wire of length 1.0 m carries a current of 48 A to the right. A second wire with a length of 0.5 m and a mass of 0.50 kg is suspend 0.30 m above the first wire. In what direction(left or right) does the current in the upper wire have to flow in order to be suspended against the force of gravity? What is the magnitude of the magnetic field (in mu T) due to the bottom wire at the location of the top wire? What the weight (in N) of the lop wire that must be balanced by the magnetic force? Report your answer without units and use only digit after the decimal. What is the magnitude of the current in the top wire? Report your answer in mega-Amps (without units) and use only on digit after the decimal.

Answer 1

length of horizontal wire = 1m ; current ( left to right ) =48A

second wire length =0.5m ; mass=0.5Kg ; height from 1st wire =0.30 m

(a) direction of current in second suspended wire = From Right to left ( because conductors carrying current in opposite directions repel each other)

(b) field at upper wire due to bottom wire

   $B = \mu _{0}i/2\pi r$    i = current in first wire r = distance of second wire from first

   $\mu _{0}$ = permeability constant = 1.26 X 10^(-6) henry/m

B = 1.26 X 10^(-6) H/m X 48 A / 2 X 3.14 X 0.3 m = 32.1 $\mu T$

(c) weight of the top wire that must be balanced = m g = 0.5 Kg X 9.8m/s^2 = 4.9 N

(d) Current in top wire to balance weight = ?

F = 4.9 N ( to balance wight of top wire) is required to be created at top wire due to bottom wire. This should be repulsive force

$F = \mu _{0}l i_{a}i_{b}/2\pi d$  where l = length of 2nd wire at top

$i_{a}$ = current in wire at bottom , $i_{b}$ = current in top wire d= distance between the wires

F = 1.26 X 10^(-6) H/m X 0.5 m X 48 A X $i_{b}$ /(2$\pi \times 0.3m$) = 4.9N

$i_{b}$ = 2 X 3.14 X 0.3 m X 49 N /( 1.26 X 10^(-6) X ).5 X 48 A) = 3.0 X 10^(6) A