9 × 10²¹ electrons flow through a cross section of the wire in one hour.
<h3>What is the relation between current and charge?</h3>
- Mathematically, current = charge / time
- In S.I. unit, Charge is written in Coulomb and time in second.
<h3>What is the amount of charge flown through a wire for one hour if it carries 0.4 A current?</h3>
- Charge= current × time
- Current= 0.4 A, time = 1 hour= 3600 s
- Charge= 0.4× 3600
= 1440 C
<h3>How many numbers of electrons present in 1440C of charge?</h3>
- One electron= 1.6 × 10^(-19) C
- So, 1440 C = 1440/1.6 × 10^(-19)
= 9 × 10²¹ electrons
Thus, we can conclude that the 9 × 10²¹ electrons flow through a cross section of the wire in one hour.
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in cgs system, plank's constant= h=6.626 x10⁻²⁶ erg s
Value of Plank's constant in SI system= 6.626 x10⁻³⁴ Js
now 1 Joule= 10⁷ ergs
so h= 6.626 x10⁻³⁴ Js (10⁷ ergs/1J)
h=6.626 x10⁻²⁷ erg s
Answer:
The voltage across the capacitor is 1.57 V.
Explanation:
Given that,
Number of turns = 10
Diameter = 1.0 cm
Resistance = 0.50 Ω
Capacitor = 1.0μ F
Magnetic field = 1.0 mT
We need to calculate the flux
Using formula of flux

Put the value into the formula


We need to calculate the induced emf
Using formula of induced emf

Put the value into the formula

Put the value of emf from ohm's law





We know that,


We need to calculate the voltage across the capacitor
Using formula of charge


Put the value into the formula


Hence, The voltage across the capacitor is 1.57 V.
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1.) True
2.) True
3.) True
4.) True
5.) True
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