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Marrrta [24]
1 year ago
5

A circular loop of radius 0.0400 m is

Physics
1 answer:
Alenkinab [10]1 year ago
4 0

Let's see

Find enclosed area of loop

\\ \rm\dashrightarrow \pi r^2

\\ \rm\dashrightarrow 3.14(0.04)^2

\\ \rm\dashrightarrow 0.005m^2

Now

\\ \rm\dashrightarrow \phi=BAcos\theta

  • We need B

\\ \rm\dashrightarrow 9.59\times 10^{-7}=B(0.005)cos75

\\ \rm\dashrightarrow 9.59\times 10^{-7}=0.0046B

\\ \rm\dashrightarrow B=2084.78\times 10^{-7}

\\ \rm\dashrightarrow B=2.08\times 10^{-4}T

\\ \rm\dashrightarrow B=0.208mT

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x(t)=A₀e^{\frac{-bt}{2m}}cos(w't+\phi)=A(t)cos(w't+\phi)

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           k=9.8 N/m

           b=210 g/s= 0.21 kg/s

           A₀=13 cm

a) A(t)=A₀/8

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let n be number of oscillations made

then, nT'=t

⇒n=\frac{24}{2.2}=11(approx)

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