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mixas84 [53]
3 years ago
12

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

Physics
1 answer:
nignag [31]3 years ago
6 0

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!

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