Solutions to Week 5 Assignment

1. We assume that we want the direction of B that produces the maximum force, i,e., perpendicular to v.

    Because the charge is positive, we point our thumb in the direction of F and our fingers in the directi0on of v. to find the direction of B, we note which way we should curl our fingers, which will be the direction of the magnetic field B.

      (a) Thumb out, fingers left, curl down.

      (b) Thumb up, fingers right, curl in.

      (c) Thumb down, fingers in, curl right.

2. The magnetic force provides the centripetal acceleration:

                        qvB = mv2/r, or mv = qBr.

    The kinetic energy of the electron is

                       KE = 1/2mv2 = 1/2(qBr)2/m

                             = 1/2[(1.6 x 10-19C)(1.15T)(8.40 x 10-3 m)]2/(1.67 x 10-27 kg)

                             = 7.15 x 10-16 J

                             = (7.15 x 10-16 J)(1.60 x 10-19 J/eV)

                             = 4.47 x 103 eV = 4.47 keV.

3. The magnetic field to the west of a wire with a current to the north will be up, with a magnitude

                     Bwire = (m0/4p)2I/r

                              = (10-7 T·m/A)2(12.0A)/(0.200m) = 1.20 x 10-5 T.

    The net downward field is

                     Bdown = BEarth sin40?- Bwire

                                = (5.0 x 10-5 T) - 1.20 x 10-5 T = 2.01 x 10-5 T.

    The northern component is BEarth = Bdowncos40?= 3.83 x 10-5 T.

    We find the magnitude from

                  B = [(Bdown)2 + (Bnorth)2]1/2

                     = [(2.01 x 10-5 T)2 + (3.83 x 10-5 T)2]1/2 =  4.3 x 10-5 T.

   We find the direction from

                tan q = Bdown/Bnorth = (2.01 x 10-5 T)/(3.83 x 10-5 T) = 0.525, or q = 28?below the horizontal.

4. Because the currents and the separations are the same, we find the force per unit length between any two wires from

                 F/L = I1(m0I2/2pd) = m0I2/2pd

                        = (4p x 10-7 T·m/A)(8.0A)2/(0.380m)

                        = 3.37 x 10-5N/m.

 The directions of the forces are shown on the diagram.

 The symmetry of the force diagrams simplifies the vector addition, so we have

                FA/L = 2(F/L)cos30º

                         = 2(3.37 x 10-5N/m)cos30º = 5.8 x 10-5N/m up.

                FB/L = F/L

                        
= 2(3.4 x 10-5N/m) cos60º = 3.4 x 10-5N/m below the line toward C.

                FC/L = F/L 

                         = 2(3.4 x 10-5N/m) cos60º
= 3.4 x 10-5N/m below the line toward B.