Question

3) In the figure below, the magnetic flux through the loop shown increases according to the equation, Q = 8t3 – 312 where 0 i

3) In the figure below, the magnetic flux through the loop shown increases according to the equation, \(\Phi=8 t^{3}-3 t^{2}\) where \(\Phi\) is in webers and \(t\) is in seconds:

a) What is the induced voltage across the resistor at \(t=2\) seconds? 

b) What is the polarity of the voltage across the resistor

c) If the area of the entire closed loop is \(14 \mathrm{~m}^{2}\), what is the magnitude of the magnetic field at \(t=2\) seconds? 

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Answer #2

a)
Magnetic flux, \phi = 8 * t3 - 2 * t2
Induced EMF, E(t) = d\phi/dt
= d/dt [8 * t3 - 2 * t2]
= 24 * t2 - 4 * t

EMF at t = 2 s,
E = 24 * 22 - 4 * 2
= 88 V
--------------------------------------
b)
The magnetic flux is increasing through the area of the loop in the out of the page direction. So the induced magnetic field must be into the page. For this, the current must be clockwise. The current through the resistor is towards left, which is same as the direction of the polarity of voltage, from right to left.

c)
\phi = B * A
B = \phi /A

\phi = 8 * t3 - 2 * t2
At t = 2 s,
\phi = 8 * 23 - 2 * 22 = 56 Wb
A = 14 m2

B = 56 / 14
= 4 T

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Answer #3

We will be due Salle know that magnitude of induced voltage across to flux chonse, dd resistor here Soi e= do و ہے۔ =) ea dt

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Answer #5

6 0-813-382 Es-de (83 312) दा at - (2412_66) --6+ (41-1) € (t=2 sec ) = 6(2)(8-1) - 84 volt Pis increasing with time EMF in d

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Answer #4

Given 0 = 8t3-3ť using faradays law, induce emf is given by a) ع Eat t=2 t=2 Eat t=2= do dt dd at o (2-1) = 446 Eat tez = [2C) Given area A = 14 m2 t 25 induced emf at &=25 is 84V as part (a) Now, we have, E = BA t Et 184x2T 11 B = (84x2 А B = 12T H

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Answer #1

3Ca) We have 여 그883 342 ㅋ 24t르 6t G 1월 dh 246tet t = 28. lel 24x4 - 6x2=940 le) - 84u eincreasing According to As the flume Q is end » B es also eincreasing is out of the

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