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Magnetism And Matter

Question
CBSEENPH12037721

A circular coil of 16 turns and radius 10 cm carrying a current of 0.75 A rests with its plane normal to an external field of magnitude 5.0 x 10–2 T. The coil is free to turn about an axis in its plane perpendicular to the field direction. When the coil is turned slightly and released, it oscillates about its stable equilibrium with a frequency of 2.0 s–1. What is the moment of inertia of the coil about its axis of rotation?

Solution

Given,
Number of turns in the coil, N = 16
Radius of the coil, r = 10 cm = 0.1 m
Current passing thropugh the coil, I = 0.75 A
External magnetic field, B = 5.0 × 10-2T
Frequency of oscillation, ν = 2.0 s-1
Magnetic moment is given by, M= NIA 

Therefore,
M = NIπr2
   = 16 × 0.75 × 227(0.1)2 
   = 0.377 JT-1

Now, using formula  ν = 12πMBI 

                 v2 = MB4π2I
   
                       I = MB4π2ν2
Putting values, we get
             
                I = 0.377 × 5.0 × 10-24×2272×22   = 1.2 × 10-4 kg m2

Tips: -

 

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