Question

calculate the no of electrons flowing through the cross section of conductor when current of 20 ma flows for 50 seconds​

Answer 1

Answer :-

6.25 × 10¹⁸ electrons are flowing through the cross section of the conductor .

Explanation :-

We have :-

→ Current (I) = 20 mA

→ Time taken (t) = 50 seconds

→ Charge of an electron (e) = 1.6 × 10⁻¹⁹ C

To find :-

→ Number of electrons (n) .

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Firstly, let's convert the given unit of current from mA to A .

⇒ 1 mA = 0.001 A

⇒ 20 mA = 20(0.001)

⇒ 0.02 A

Now, we know that :-

I = Q/t

⇒ I = ne/t

⇒ 0.02 = (n × 1.6 × 10⁻¹⁹)/50

⇒ 0.02(50) = 1.6 × 10⁻¹⁹ n

⇒ 1 = 1.6 × 10⁻¹⁹ n

⇒ n = 1/(1.6 × 10⁻¹⁹)

⇒ n = 1/1.6 × 1/10⁻¹⁹

⇒ n = 0.625 × 10¹⁹

⇒ n = 6.25 × 10¹⁸

Answer 2

• $\boxed{\sf{\purple{No\:of\:electrons\:=\:\bf{6.25 \:\times\:10^{18}}}}}$

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Explanation :

$\underline{\underline{\bf{\green{Given\::-}}}}$

• Current, I = 20 mA
• Time taken, t = 50 seconds
• Charge of an electron, e = 1.6 × 10^-19

$\underline{\underline{\bf{\green{To\:Find\::-}}}}$

• Number of electrons = ?

$\underline{\underline{\bf{\green{Solution\::-}}}}$

★ Converting unit of current into A :-

✏ㅤㅤㅤ1 mA = 0.001 A

✏ㅤㅤㅤ20 mA = 20 × 0.001

✏ㅤㅤㅤ20 mA = 0.02 A

★ Using formula :-

$\qquad\red\bigstar\:{\underline{\boxed{\bf{I = \dfrac{Q}{t}}}}}$

$\qquad\leadsto\quad\sf I = \dfrac{ne}{t}$

$\qquad\leadsto\quad\sf 0.02 = \dfrac{n\:\times\:1.6\:\times\:10^{-19}}{50}$

$\qquad\leadsto\quad\sf 0.02 \:\times\:50 = (1.6\:\times\:10^{-19})n$

$\qquad\leadsto\quad\sf \dfrac{1}{1.6\:\times\:10^{-19}} = n$

$\qquad\leadsto\quad\sf 0.625\:\times\:10^{19} = n$

$\qquad\leadsto\quad\bf{ n = \red{6.25\:\times\:10^{18}}}$

$\small\therefore\:{\underline{\sf{Hence,\:number\:of\:electrons\:=\:\bf{6.25\:\times\:10^{18}}}}}$

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