A square coil of side 16 cm has 200 turns and rotates in the uniform magnetic field of magnitude 0.05 T. If the peak emf is 12 V, what is the angular velocity of the cell? (Ans: 47 rad s⁻¹). 



Data Given:

Length of the side of square = `\l` = 16 cm = 0.16 m

Thus area is = A = 0.16 m ｘ 0.16 m = 0.0256 m² = 2.56 ｘ10⁻² m²

Number of turns = N = 200 

Magnetic field strength = B = 0.05 T

Peak value emf = `\Ɛ_0` = 12 V



To Find:


Angular velocity = ധ = ?


Solution:

The relation for the Peak value emf  `\Ɛ_0` is given 

`\Ɛ_0`  = B ധ N A

or

ധ = `\frac {Ɛ_0}{BNA}`


Putting the corresponding values 


ധ = `\frac {12 V}{0.05 T ｘ 200 ｘ 2.56 ｘ10⁻² m²}`


ധ = `\frac {12 V}{25.6 ｘ10⁻² Tm²}`

ധ = 0.46875ｘ10² rad s⁻¹

or

ധ = 46.875 rad s⁻¹

or

ധ = 47 rad s⁻¹  -------------Ans.






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