# (a)Derive an expression for induced emf developed when a coil of N turns, and area of cross -section, A is rotated at a constant angular speed $\omega$ in uiform magnetic field B.(b) A wheel with 100 metallic spokes each 0.5m long is rotated with speed of 120rev/min in a plane normal to the horizontal component of Earth's magnetic field. If the resultant magnetic field at that plane is $4\times 10^{-4}T$ and angle of dip at place is $30^{0}$, find the emf induced between the axle and the rim of the wheel.

(a) Deriving the expression for emf when a coil of N turn, and area of cross-section A, is rotated at a constant angular speed $\omega$ in a uniform magnetic field B.

Let 'A' be the area of each turn of the coil.

'N' be the number of turns in the coil

$\overrightarrow{B}$ strength of the magnetic field

$\phi = N \left ( \overrightarrow{B}.\overrightarrow{A} \right )=NBA \cos \omega t$   ......(i)

where $\omega$ is angular velocity of the coil.

As the coil rotates, angle $\theta$ change, therefore magnetic flux $\phi$ linked with the coil changes and hence, an emf is induced in the coil.

$e = \frac{-d\phi}{dt} = \frac{-d}{dt}(NAB \cos \omega t)$

$e = -NAB \frac{d}{dt}(\cos \omega t)$

$e = -NAB (-\sin \omega t)\omega$

therefore, $e = NAB (\sin \omega t)\omega$

(b)  We have given the values:

length of metallic spokes (l) =0.5m

Speed or velocity (v) = 120rpm = 2rps

Hence,$\omega = 2\pi \nu = 4\pi rad/s$

$B = 4 \times 10^{-4}T$

$\delta = 30^{\circ}$(angle of dip)

Now, $B_{H}=4\times 10^{-4}\times \frac{\sqrt{3}}{2}$

Where $B_{H}$ is  magnetic field

So, $B_{H}= 2\sqrt{3}\times 10^{-4}T$

$\varepsilon =\frac{1}{2}B\omega l^{2}$

$\varepsilon =\frac{1}{2}\times 2\sqrt{3}\times 10^{-4}\times 4\pi \times (0.5)^{2}$

$\varepsilon = 5.4\times 10^{-4}Volt$

Hence, the emf induced between the axle and the rim of the will is $5.4\times 10^{-4}Volt$.

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