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Home / Doubts and Solutions/JEE/Physics

Ques13

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Asked by rbhatt16 19th July 2018, 9:28 AM
Answered by Expert
Answer:
If a magnetic dipole having moment M be rotated through angle θ from equilibrium position in a uniform magnetic field B,
workdone on it is W = MB(1-cosθ)
 
This work is stored in the system in the form of Energy.
When system is released, dipole starts to rotate to occupy equilibrium position and the energy converts into kinetic energy.
Kinetic energy of the system is maximum when stored energy is completely released.
 
Magnetic Induction, at centre due to current in larger coil is begin mathsize 12px style B space equals space fraction numerator mu subscript o I over denominator 2 R end fraction end style
magnetic dipole moment of smaller coil is i×π×r2  
 
Initially, planes of two coils are perpendicular, therfore θ is 90º. 
Energy of the system is U = ( i×π×r2  )×B×(1-cos 90) = begin mathsize 12px style equals space left parenthesis i pi r squared right parenthesis fraction numerator mu subscript o I over denominator 2 R end fraction end style
When coils are released, both the coils start to rotate about their common diameter and
their kinetic energies are maximum when they become coplanar.
 
Moment of inertia of larger coil about axis of rotation is I1 = (1/2)MR2 and smaller coil is I2 = (1/2)mr2 .
 
Since two coils rotate due to their mutual interaction only, therfore if one coil rotates clockwise then other rotates in anticlockwise
 
Let angular velocities of larger and smaller coil be numerically equal to ω1 and ω2  respectively when they become coplanar.
 
According to law of conservation of angular momentum,  I1 ω1  = I2 ω2 ................... (1)
and according to law of conservation of energy, begin mathsize 12px style 1 half I subscript 1 omega subscript 1 superscript 2 space plus space fraction numerator begin display style 1 end style over denominator begin display style 2 end style end fraction I subscript 2 omega subscript 2 superscript 2 space equals space U end style ...................(2)
from (1) and (2), we have begin mathsize 12px style 1 half I subscript 2 omega subscript 2 superscript 2 space equals space fraction numerator U cross times I subscript 1 over denominator open parentheses I subscript 1 plus I subscript 2 close parentheses end fraction space equals space fraction numerator mu subscript o pi cross times I cross times i cross times M cross times R cross times r squared over denominator 2 open parentheses M R squared plus m r squared close parentheses end fraction end style
Answered by Expert 3rd August 2018, 12:30 PM
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