Question

An inductor has inductance of 0.260 H and carries a current that is decreasing at a uniform rate of 18.0 mA/s.

Answer 1

Answer:

The self-induced emf in this inductor is 4.68 mV.

Explanation:

The emf in the inductor is given by:

$\epsilon = -L\frac{dI}{dt}$

Where:

dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)

L: is the inductance = 0.260 H

So, the emf is:

$\epsilon = -L\frac{dI}{dt} = -0.260 H*(-18.0 \cdot 10^{-3} A/s) = 4.68 \cdot 10^{-3} V$

Therefore, the self-induced emf in this inductor is 4.68 mV.  

I hope it helps you!