91影视

1. Turns of the coil N
2. Length of the rectangular coil is a
3. Width of the rectangular coil is b
4. Frequency of rotation,$\mathrm{f}=60\frac{\mathrm{rev}}{\mathrm{s}}$
5. Induced emf $\mathrm{蔚}=150\mathrm{V}$
6. Magnetic field, B = 0.500 T

Use the expression for magnetic flux in Faraday鈥檚 law and deriving it, prove the required equation. By using the same proved equation, calculate value of Nab.

Faraday's law of electromagnetic inductionstates, Whenever a conductor is placed in a varying magnetic field, an electromotive force is induced in it.

Formulae are as follow:

Faraday's law,

$蔚=\frac{d蠁}{dt}$

Magnetic flux, $蠁=蠒BdA$

Where$\mathrm{桅}$ is magnetic flux, B is a magnetic field, A is an area,饾渶 is emf.

It is given that, a coil of conducting wire is placed in a magnetic field B.

The number of turns in a coil are N. The coil is rotated at frequency f.

Let the area of each turn be A.

Angular velocity of the coil is $蝇=2蟿蟿f...................................................\left(1\right)$

The coil is rotated by an angle $胃$, in time t then $胃=蝇t..........................................................\left(2\right)$

Magnetic flux passing through the coil is given by,

$$\begin{gathered} 蠁=蠒BdA=蠁BdA\cos 胃 \\ 蠒dA=AreaofNturns=NA \\ 蠁=NAB\cos 胃 \end{gathered}$$

From equation 1,

$蠁=NAB\cos 蝇t.........................................................\left(3\right)$

Using this in the emf induced in a coil,

$$\begin{gathered} 蔚=\frac{d蠁}{dt} \\ 蔚=\frac{d}{dt}\left[NAB\cos 蝇t\right] \\ 蔚=-NAB[-蝇\sin 蝇t] \\ 蔚=NAB蝇\sin 蝇t \end{gathered}$$

From equation 1,

$蔚=NAB2蟿蟿f\sin 2蟿蟿ft$

Since, it is a rectangular coil, area$A=length脳width=a脳b$

$蔚=Nab2蟿蟿f\sin 2蟿蟿ft$

This is an expression for the induced emf in the coil at any instant t.

When $\sin 2蟿蟿ft$= 1,the emf is maximum.

$$\begin{gathered} {蔚}_{max}={蔚}_0=NabB2蟿蟿f.........................................................\left(3\right) \\ 蔚={蔚}_0\sin 2蟿蟿ft \end{gathered}$$

Hence, $蔚=2蟿蟿fNabB\sin 2蟿蟿ft={蔚}_0\sin 2蟿蟿ft$is proved.

It is given that,

Frequency of rotation,$\mathrm{f}=60\frac{\mathrm{rev}}{\mathrm{s}}$

Induced emf $\mathrm{蔚}=150\mathrm{V}$

Magnetic field, B = 0.550 T

And now find Nab.

At maximum emf,$\sin 2蟿蟿ft=1$.

Using all this in equation 3,

$$\begin{gathered} \mathrm{蔚}=\mathrm{NabB}2\mathrm{蟿蟿蹿} \\ \mathrm{Nab}=\frac{\mathrm{蔚}}{\mathrm{B}2\mathrm{蟿蟿蹿}} \\ \mathrm{Nab}=\frac{150}{0.5脳2脳3脳3.14脳60} \\ \mathrm{Nab}=0.7961{\mathrm{m}}^2 \end{gathered}$$

Hence, the value of Nab is $0.7961{\mathrm{m}}^2$.

Therefore, use the expression for magnetic flux in Faraday鈥檚 law and by deriving it, prove the required equation. By using the same equation, calculate the value of Nab.