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When a coil is carrying a current of 25.0 A that is increasing at 145 A/s the induced emf in the coil has magnitude 3.70 mV.
A. Find the self-inductance of the coil.
B. Find the magnitude of the induced emf at t = 2.00 s if the current is i(t) = (225 A/s)t2.

Respuesta :

Answer:

a) 25.5 µH

b) 22.95 mV

Explanation:

Induced emf in a inductor is given by

E = L * di/dt, where

E is the voltage of the circuit

L is the inductance of the circuit

di/dt if the rate of inductance

A

So we have

0.0037 = L * 145

L = 0.0037 / 145

L = 0.0000255

L = 25.5 µH

B

i(t) = 225t²

Recall that

E = L * di/dt, so that

E = 25.5 µH * |225t²|

Differentiating with respect to t, we have

E = 25.5 * 2 * 225t

E = 25.5 * 450t

Solving for t = 2,we get

E = 25.5 * 450(2)

E = 25.5 * 900

E = 22950 µV or

E = 22.95 mV

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