42.
(a) The force on the left wire points left, that on the right wire
points right, and both have magnitude
mu0 a+b
F = ----- I1 I2 ln (---)
2 pi a
(the two forces cancelling)
Fbot= [mu0 / (2 pi)] I1 I2 L / (a+b) down
Ftop= [mu0 / (2 pi)] I1 I2 L / a up
Fnet= [mu0 / (2 pi)] I1 I2 L b / [a (a + b)] up
43. Proof.
44. Proof
45. Proof
Last revised 1998/04/17 |