(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.

 

 

 

 
 
 
 
 

Answers (1)

(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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