1. Multiloop circuit In Fig. 27-71, R1 = 5.76 , R2 = 18.7 , and the ideal battery has emf = 13.8 V. (a) What is the size of current i1? http://img253.imageshack.us/img253/9926/fig2771dw3.gif Relevant equations i = V/R The attempt at a solution I redrew the 3 resistors (R2s) as parallel resistors. So i1 would be in the middle. Then all the resistances would have the same voltage 13.8V. So i1 = V/R2 = 13.8V/18.7 = 1.59 A. It's wrong though. 2. Magnetic Force on a Current-Carrying Wire A wire 62.9 cm long carries a 0.480 A current in the positive direction of an x axis through a magnetic field with an x component of zero, a y component of 0.000230 T, and a z component of 0.0120 T. Find the magnitudes of x, y, and z components of the force on the wire. Relevant equations Magnitude of the force on a current carrying wire = (current)*(length)*(magnetic field)*(sin (angle between the directions of field and the length)) The attempt at a solution The length of wire is along the x axis, so the angle between the wire and the x axis = 0. The angle between the x axis and y axis = 90 degrees, between x and z is also 90 degrees: force magnitude of x component = 0 force magnitude of y component = 0.480A x 0.629m x 0.000230T x sin 90 force magnitude of z component = 0.480A x 0.629m x 0.0120T x sin 90 Only the first one's right. 3. Ampere's Law In Fig. 29-63, a long circular pipe with outside radius R = 2.65 cm carries a (uniformly distributed) current i = 9.16 mA into the page. A wire runs parallel to the pipe at a distance of 3.00R from center to center. Find the magnitude of the current in the wire in milliamperes such that the ratio of the magnitude of the net magnetic field at point P to the magnitude of the net magnetic field at the center of the pipe is 3.93, but it has the opposite direction. http://img442.imageshack.us/img442/8181/fig2963ho7.gif Relevant equations Magnetic field at distance r outside long straight wire with current = (μ0 x i)/(2 x pi x r) Magnetic field around Amperian loop = μ0 x current enclosed The attempt at a solution (Magnetic Field at pt P) / (Magnetic Field at the center of the pipe) = 3.93 so: (μ0 x i)/(2 x pi x R) = 3.93 x (μ0 x current enclosed) ...i = 3.93 x current enclosed x 2 x pi x R = 3.93 x 9.16 mA x 2pi x 0.0265m = 0.00599 (mA)(m) The units should only have mA so the answer is wrong.