Giancoli 5th Ed.

Hwk 8  (Due 10/21/22)        10 point hwk
Ch. 26 P. 1,2,5,7,12,33
Ch. 27 P. 3-5,10,13,17,21,28,33,39,49

Ch. 26 Problems:
1. a) Vab=5.91 V
   b) Vab=5.99 V
2. Vab = 4(emf - Ir) = IR
      r = [emf - (IR/4)h]/I  = 0.2 ohm
5. V=IR_2  where I=emf/(R_1+R_2)
   V=8.6 V
7. R_eq = (5/100)^{-1} = 20 ohms = n(10 ohms)
   so n=2
12. a) V = IR = (Vtot/8R)R = Vtot/8 = 120 / 8 = 15 V
    b) R=V_tot/8I = 23 ohms  and P=I^2R =9.8 W
17. a) R_eq = 588 ohms
    b) I_tot = V/R_eq = .0851 A
       I = .0624 A through left-most vertical resistor
33.  I2 = 0.27 A, left
     I1 = 0.60 A, left
     (also, I3 = 0.8762 A)


Ch. 27 Problems: 
3. a) F/l = IBsin(theta) = 6.7 N/m
   b) F/l = IBsin(theta) = 3.8 N/m
4. I = F_max/(lB) = 1.86 A
5. F=IlBsin(68)=1.7N
10. Fmax = IlB, B = 0.458 T
13. B = 0.34 i + 0.17 k  T
17. Fmax = qvB = 8.1e-14 N
    direction = North
21. a) no force
    b) downward
    c) upward
    d) inward, into the paper
    e) to the left
    f) to the left
28. F=qvxB = 3.2e-15 N k
33. a) charge must be negative
    b) F_max = qvB=mv^2/r, so p=mv=qBr
                              p=qB(d^2+l^2)/2d
39. torque=NLABsin(theta) = 0.359mN
49. a) F_E = F_B , qE=qvB, v=E/B=5.85e+6 m/s.


Please check               Tot= 10 pts
Ch. 27 Prob.  13 28