A circular wire loop of radius 1 , cm carries | vedclass

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Question

(C) Radius of the loop, $r = 1 \, cm = 0.01 \, m$. Total charge on the loop, $Q = 1 	imes 10^{-6} \, C$. Charge per unit length, $\lambda = \frac{Q}{

Vedclass · Accepted Answer

$9 	imes 10^3 \, N/C$

Vedclass · Answer

(C) Radius of the loop, $r = 1 \, cm = 0.01 \, m$. Total charge on the loop, $Q = 1 	imes 10^{-6} \, C$. Charge per unit length, $\lambda = \frac{Q}{2 \pi r} = \frac{10^{-6}}{2 \pi 	imes 0.01} = \frac{10^{-4}}{2 \pi} \, C/m$. The length of the cut-off part is $l' = 0.01 \% \, 	ext{of} \, 2 \pi r = \frac{0.01}{100} 	imes 2 \pi r = 2 \pi 	imes 10^{-6} \, m$. The charge on the cut-off part is $Q' = l' \lambda = (2 \pi 	imes 10^{-6}) 	imes \frac{10^{-4}}{2 \pi} = 10^{-10} \, C$. For a complete loop, the electric field at the centre is zero. If a small part $l'$ is removed, the electric field at the centre due to the remaining part is equal in magnitude to the electric field that would have been produced by the removed part $l'$ at the centre. Treating the small cut-off part as a point charge $Q'$ at distance $r$ from the centre, the electric field is $E = \frac{k Q'}{r^2}$. $E = \frac{9 	imes 10^9 	imes 10^{-10}}{(0.01)^2} = \frac{0.9}{10^{-4}} = 9000 \, N/C = 9 	imes 10^3 \, N/C$.

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