- #1
Meow12
- 45
- 20
- Homework Statement
- The inductor in the figure has inductance 0.260 H and carries a current in the direction shown that is decreasing at a uniform rate, ##\displaystyle\frac{di}{dt}## = -0.0180 A/s.
(a) Find the self-induced emf.
(b) Which end of the inductor, a or b, is at a higher potential?
- Relevant Equations
- ##\displaystyle\mathcal E=-L\frac{di}{dt}##
(a) ##\displaystyle\mathcal E=-L\frac{di}{dt}## = -0.260 x -0.0180 = 0.00468 V = 4.68 mV
(b) Current always flows from higher potential to lower potential. So, b is at a higher potential.
My textbook says that my answer for part (a) is correct but part (b) is wrong. Please help!