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Yuliya22 [10]
3 years ago
15

A physics student notices that the current in a coil of conducting wire goes from i1 = 0.200 A to i2 = 1.50 A in a time interval

of Δt = 0.300 s. Assuming the coil's inductance is L = 2.00 mH, what is the magnitude of the average induced emf (in mV) in the coil for this time interval?
Physics
1 answer:
Afina-wow [57]3 years ago
7 0

To solve the problem it is necessary to apply the concepts related to Voltage in an Inductor. By general definition the voltage in in an inductor is defined as

\epsilon = L \frac{di}{dt}

Where,

\epsilon =Voltage induced (emf)

L = Inductance

di = Rate of change of current flow

dt = rate of change of time

Our values are given by

L = 2mH = 2*10^{-3} H

\Delta i = i_2-i_1 = 1.5A-0.2A = 1.3A

\Delta t = 0.3s

Replacing

\epsilon = 2*10^{-3}*\frac{1.3}{0.3}

\epsilon = 8.6*10^{-3}V

\epsilon = 8.6mV

Therefore the magnitude of average induced emf is 8.6mV

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      Time elapsed before the bullet hits the ground is 0.553 seconds.

b) Consider the horizontal motion of bullet

We have equation of motion s = ut + 0.5 at²

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                      s = ut + 0.5 at²

                      s = 200 x 0.553 + 0.5 x 0 x 0.553²

                      s = 110.6 m

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