121k views
4 votes
A coil has an inductance of 2.00 mH, and the current in it changes from 0.200 A to 1.50 A in a time interval of 0.150 s. Find the magnitude of the average induced emf in the coil during this time interval.

User Hrickards
by
5.4k points

2 Answers

0 votes

Answer:

0.0173 V

Step-by-step explanation:

The Emf induced in a coil is given as

E = Ldi/dt......................... Equation 1

Where E = Emf induced in a coil, L = Inductance of the coil, di = change in current, dt = time interval/change in time.

Given: L = 2.0 mH = 2×10⁻³ H, di = 1.5-0.2 = 1.3 A, dt = 0.15 s

Substitute into equation 1

E = 2×10⁻³(1.3/0.15)

E = 1.73×10⁻² V

E = 0.0173 V

Hence the emf induced in the coil = 0.0173 V

User Frankegoesdown
by
5.4k points
3 votes

Answer:

0.0173 V

Step-by-step explanation:

PARAMETERS GIVEN:

Inductance, L = 2 mH = 0.002 H

Time interval, dt = 0.15 s

Change in current, dI = 1.5 - 0.2 = 1.3 A

The magnitude of the induced EMF is given as:

EMF = | -L*dI/dt |

EMF = (0.002 * 1.3) / 0.15

EMF = 0.0173 V

User Sho Gondo
by
6.8k points