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Asked by jeeasp7676 | 18 Aug, 2023, 12:32: PM
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Let the square loop is on y-z plane and direction of magnetic field B is in -x axis direction as shown in figure. Axis of rotation OO' is parallel to z-axis
 
If small angular displacement θ is made to the loop , then the restoring torque τ acting on the loop due to  begin mathsize 14px style stack i space l space with rightwards arrow on top space cross times space B with rightwards arrow on top space end styleforce ( Refer top view figure )
acting on the vertical side of loop is
 
begin mathsize 14px style tau space equals negative F space cross times space l space cross times sin theta space equals negative space left parenthesis space i space l squared space B space right parenthesis space sin theta end style .............................. (1)
where i is the current passing through loop and l is the side length of loop.
 
If I is the moment of inertia , then the torque begin mathsize 14px style tau space equals space I space cross times space alpha end style , where  begin mathsize 14px style alpha space equals space fraction numerator d squared theta over denominator d t squared end fraction end styleis angular acceleration
Hence , eqn.(1) can be written using small angle approximation begin mathsize 14px style sin theta almost equal to theta end style as
begin mathsize 14px style I space cross times fraction numerator d squared theta over denominator d t squared end fraction space equals space minus left parenthesis i space l squared B space right parenthesis space theta end style
 
begin mathsize 14px style fraction numerator d squared theta over denominator d t squared end fraction space equals space minus space fraction numerator left parenthesis i space l squared B space right parenthesis over denominator I end fraction space theta end style
 
If we compare above expression with the standard SHM equation begin mathsize 14px style fraction numerator d squared theta over denominator d t squared end fraction space equals space minus space omega squared space theta end style,
then we get angular frequency of oscillatory motion of coil as
 
begin mathsize 14px style omega space equals space square root of fraction numerator i space l squared B over denominator I end fraction end root end style........................... (2)
 
 Moment of inertia of loop about the given axis of rotation is
 
begin mathsize 14px style I space equals space 2 space cross times m over 4 cross times fraction numerator l squared over denominator 3 space end fraction space plus space 2 space cross times m over 4 cross times l squared over 4 space equals space 7 over 24 m l squared end style
 
 By substituting moment of inertia begin mathsize 14px style I end style in eqn. (2), we get
 
begin mathsize 14px style omega space equals space square root of 24 over 7 fraction numerator i space B over denominator m end fraction end root end style
 
Angular frequency ω and period T of oscillation are related as
 
begin mathsize 14px style T space equals space fraction numerator 2 pi over denominator omega end fraction end style
 
Hence period of oscillation T is
 
begin mathsize 14px style T space equals space 2 pi space square root of 7 over 24 fraction numerator m over denominator i space B end fraction end root space end style
 
Answered by Thiyagarajan K | 19 Aug, 2023, 01:08: PM
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answer
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in this quesn correct answer are give 1,2,3. but in 2,3 magnetic field is a vector quantity. so to make it same at 2 points we need direction and magnitude to be same. But in case of straight wire magnitude can be same at a radius but the direction cant be same. So how 2 and 3 are correct AND 4 should be correct because in formula B is inversely prop to r are the answers given wrong ?
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