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(D) Let the total charge on the loop be $Q = 10^{-5} \ C$ and the radius be $a = 10 \ cm = 0.1 \ m$. The linear charge density is $\lambda = \frac{Q}{

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(D) Let the total charge on the loop be $Q = 10^{-5} \ C$ and the radius be $a = 10 \ cm = 0.1 \ m$. The linear charge density is $\lambda = \frac{Q}{2 \pi a}$. The charge on the small cut-off element of length $dl = 3.14 	imes 10^{-6} \ m$ is $dq = \lambda dl = \frac{Q dl}{2 \pi a}$. Initially,the electric field at the center due to the complete loop is zero. If $E_{	ext{rem}}$ is the field due to the remaining wire and $E_{dl}$ is the field due to the cut-off element,then $E_{	ext{rem}} + E_{dl} = 0$,which implies $|E_{	ext{rem}}| = |E_{dl}|$. The electric field due to the small element $dq$ at the center is $E_{dl} = \frac{k dq}{a^2} = \frac{k Q dl}{a^2 (2 \pi a)} = \frac{k Q dl}{2 \pi a^3}$. Substituting the values: $E_{dl} = \frac{(9 	imes 10^9) 	imes (10^{-5}) 	imes (3.14 	imes 10^{-6})}{2 	imes \pi 	imes (0.1)^3}$ Since $2 \pi \approx 2 	imes 3.14 = 6.28$,we have $3.14 / (2 \pi) = 0.5$. $E_{dl} = \frac{9 	imes 10^9 	imes 10^{-5} 	imes 10^{-6} 	imes 0.5}{0.001} = \frac{9 	imes 10^{-2} 	imes 0.5}{10^{-3}} = 4.5 	imes 10^1 = 45 \ N \ C^{-1}$.

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