If the flux of magnetic induction through a coil of resistance R and having n turns changes from -1 to -2- then the magnitude of the charge that passes through the coil is

Induced emf is |e| = n ΔΦ/Δ t, Now Δ q = 1 Δ t = e/RΔ t = n ΔΦ/R Δ t×Δ t = n ΔΦ/R = n(Φ_2 Φ_1)/R ...

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Standard XII

Physics

Alternating Current

If the flux o...

Question

If the flux of magnetic induction through a coil of resistance R and having n turns changes from $Φ_{1}$ to $Φ_{2}$ , then the magnitude of the charge that passes through the coil is

$\frac{Φ_{2} - Φ_{1}}{R}$

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$n \frac{Φ_{2} - Φ_{1}}{R}$

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$\frac{Φ_{2} - Φ_{1}}{n R}$

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$\frac{n R}{Φ_{2} - Φ_{1}}$

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Solution

The correct option is B
$n \frac{Φ_{2} - Φ_{1}}{R}$

Induced emf is $| e | = n \frac{Δ Φ}{Δ t}$ , Now
$Δ q = 1 Δ t = \frac{e}{R} Δ t = \frac{n Δ Φ}{R Δ t} \times Δ t = \frac{n Δ Φ}{R} = \frac{n (Φ_{2} - Φ_{1})}{R}$

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If the flux of magnetic induction through a coil of resistance R and having n turns changes from Φ1 to Φ2, then the magnitude of the charge that passes through the coil is

The correct option is B nΦ2−Φ1R Induced emf is |e|=nΔΦΔt, Now Δq=1 Δt=eRΔt=nΔΦRΔt×Δt=nΔΦR=n(Φ2−Φ1)R

If the flux of magnetic induction through a coil of resistance R and having n turns changes from $Φ_{1}$ to $Φ_{2}$ , then the magnitude of the charge that passes through the coil is

The correct option is B
$n \frac{Φ_{2} - Φ_{1}}{R}$

Induced emf is $| e | = n \frac{Δ Φ}{Δ t}$ , Now
$Δ q = 1 Δ t = \frac{e}{R} Δ t = \frac{n Δ Φ}{R Δ t} \times Δ t = \frac{n Δ Φ}{R} = \frac{n (Φ_{2} - Φ_{1})}{R}$