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.

Required:
Find the self-induced emf in this inductor.

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!

Answer 2

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

Calculation of the emf in the inductor:

Here

E = -LdI/dt

here

dI/dt represents the the decreasing current's rate change = -18.0 mA/s

the minus sign is due to the current is decreasing

L represents the inductance = 0.260 H

Now the emf should be

= -0260H * (-18.0.10^-3)

=  4.68 mv

hence, The self-induced emf in this inductor is 4.68 mv.

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