Suppose the equilibrium separation of the rods is r 3 mm th

Suppose the equilibrium separation of the rods is r = 3 mm, the rods are L = 10 cm long, and the current is I = 5 A. Calculate the field due to one rod at the location of the other (using equation I). Calculate the force between the rods (using equations 4 or 5). How much mass is required to keep the rods at their equilibrium position (using equation 6)? If the distance of r_meas = 3 mm was incorrectly measured and the actual distance is r = 3.5 mm. what the actual force? Calculate the percentage uncertainty in measured distance using |r - r_meas|/r. Calculate the percentage uncertainty in the force.

Solution

1.

Magnetic field due to one rod at the location of the other is given by

B =u_oI/2pir =(4pi*10^-7)*5/2pi*(3*10^-3)

B=3.33*10^-4 T or 0.333 mT

b)

Force between the rods

F=BIL =3.33*10^-4*5*0.1

F=1.67*10^-4 N or 0.167 mN

c)

Mass of the rods

m=F/g =1.67*10^-4/9.8

m=1.7*10^-5 kg or 0.017 g

d)

Actual magnetic field

B=(4pi*10^-7)*5/2pi*(3.5*10^-3)

B=2.857*10^-4 T

Magnetic force

F=(2.857*10^-3)*5*0.1

F=1.43*10^-4 N

e)

% uncertainity =(|3-3.5|/3)*100 =16.67 %

f)

% uncertainity =[|1.67-1.43|*10^-4/(1.67*10^-4)]*100

% uncertainity =14.37 %