Question

The current through an inductor of inductance L is given by I(t) = Imax sin(ωt).

Answer 1

Answer:

(a) $emf_L=-LI_{max}\omega cos(\omega t)$

(b) neither increasing or decreasing

(c) opposite to the flow of charge carriers

Explanation:

The current through an inductor of inductance L is given by:

$I(t)=I_{max}sin(\omega t)$   (1)

(a) The induced emf is given by the following formula

$emf_L=-L\frac{dI}{dt}$    (2)

You derivative the expression (1) in the expression (2):

$emf_L=-L\frac{d}{dt}(I_{max}sin(\omega t))\\\\emf_L=-LI_{max}\omega cos(\omega t)$

(b) At t=0 the current is zero

(c) At t = 0 the emf is:

$emf_L=-\omega LI_{max}$

w, L and Imax have positive values, then the emf is negative. Hence, the induced emf is opposite to the flow of the charge carriers.

(d) read the text carefully