QUIZ 4

NAME___________

 

  1. Find the currents I1, I2, and I3 in the circuit shown in Figure

 

Solution. Applying Kirchhoff’s  point rule: At point c: I1 + I2 = I3

Loop Rule:

Loop abcda:

10V – (6W)I1 –    (2W)I3 = 0

Loop befcb:

14V + 10V –(6W)I1 + (4W)I2 = 0

Solving the three equations, we get

I1 = 2A     I2 = -3A,     I3 = -1A

 

 

 

 

 

  1. A wire having a mass per unit length of 0.50g/cm carries a 2.0-A current horizontally to the south. What are the direction and magnitude of the minimum magnetic field needed to lift this wire vertically upward? Hint: Assume that the length of the wire is l.

Assume that the length of the wire is l.

The weight of the wire is W = llg.

The minimum force to lift this wire vertically upward is F = -W

F = IlBsin90°

llg = IlB

(0.50 x 10-3kg/10-2m)l(9.80m/s2) =(2.0A)lB, which gives

B = 0.245T

The direction is to the east.