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A circular wire loop of radius LaTeX: RR lies in the xy-plane with the z-axis running through its center. There is initially no magnetic fie
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A circular wire loop of radius LaTeX: RR lies in the xy-plane with the z-axis running through its center. There is initially no magnetic field present. At LaTeX: t=0t = 0 a magnetic field given by LaTeX: \vec{B}=Ce^t\hat{x}+Dt^2\hat{z}B → = C e t x ^ + D t 2 z ^ (where LaTeX: CC and LaTeX: DD are constants) is turned on. Note that this means that the field magnitude only depends on time, and that the field initially points along LaTeX: \hat{x}x ^. What is the magnitude of the emf induced in the current loop?
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Physics
4 years
2021-07-13T08:14:51+00:00
2021-07-13T08:14:51+00:00 1 Answers
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Answer:
Explanation:
Given a circular loop of radius R
r = R
Note: the radius lies in the xy plane
Area is given as
A = πr² = πR²
At t = 0, no magnetic field B=0
The magnetic field is given as a function of time
B = C•exp(t) •i + D•t² •k
Where C and D are constant
We want to find the magnitude of EMF in the circular loop.
EMF is given as
ε = – N•dΦ/dt
Where,
N is number of turn and in this case we will assume N = 1.
Φ is magnetic flux and it is given as
Φ = BA
ε = – N•d(BA)/dt
Where A is a constant, then we have
ε = – N•A•dB/dt
B = C•exp(t) •i + D•t² •k
dB/dt = C•exp(t) •i + 2D•t •k
Then,
ε = – N•A•dB/dt
ε = – 1•πR²•(C•exp(t) •i + 2D•t •k)
ε = -πR²•(C•exp(t) •i + 2D•t •k)
So, let find the magnitude of EMF
Generally finding magnitude of two vectors R = a•i + b•j
Then, |R| = √a² + b²
So, applying this we have,
ε = πR² (√(C²•exp(2t) + 4D²t²))
From the given magnetic field, we are given that,
B = 0 at t = 0
B = C•exp(t) •i + D•t² •k
B = 0 = C•exp(0) •i + D•0² •k
0 = C
Then, C = 0.
So, substituting this into the EMF.
ε = πR² (√(0²•exp(2t) + 4D²t²))
ε = πR² (√4D²t²)
ε = πR² × 2Dt
ε = 2πDR²t
So, the EMF is also a function of time
ε = 2πDR²t