A thin flexible wire of length L is connected | vedclass

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(C) Consider a small element of the wire of length $dl = R d	heta$ subtending an angle $d	heta$ at the center of the circular arc.The magnetic force

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$\frac{IBL}{2 \pi}$

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(C) Consider a small element of the wire of length $dl = R d	heta$ subtending an angle $d	heta$ at the center of the circular arc.The magnetic force on this element is $dF = I (dl) B = I (R d	heta) B$,which acts radially outward.The tension $T$ in the wire acts at both ends of this element. The net radial force due to tension is $2 T \sin(\frac{d	heta}{2}) \approx T d	heta$ for small $d	heta$.Equating the radial magnetic force to the radial component of tension:$T d	heta = I B R d	heta$$T = I B R$Since the total length of the wire is $L$,and assuming it forms a complete circle (or nearly so),$L = 2 \pi R$,so $R = \frac{L}{2 \pi}$.Substituting $R$ into the tension equation:$T = I B (\frac{L}{2 \pi}) = \frac{IBL}{2 \pi}$

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