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Question

Physics Question on Electromagnetic induction

A long solenoid of diameter 0.1m0.1\, m has 2×1042 \times 10^4 turns per meter. At the centre of the solenoid, a coil of 100100 turns and radius 0.01m0.01\, m is placed with its axis coinciding with the solenoid axis. The current in the solenoid reduces at a constant rate to 0A0\,A from 4A4\, A in 0.05s0.05 \,s. If the resistance of the coil is 10π2Ω10\pi^2 \Omega . the total charge flowing through the coil during this time is :

A

16μC16 \, \mu C

B

32μC32\, \mu C

C

16πμC16 \, \pi \, \mu C

D

32πμC32 \, \pi \, \mu C

Answer

32μC32\, \mu C

Explanation

Solution

Sol. ε=Ndϕdt\varepsilon=-N \frac{d \phi}{d t}
εR=NRdϕdt\left|\frac{\varepsilon}{R}\right|=\frac{N}{R} \frac{d \phi}{d t}
dq=NRdϕd q=\frac{N}{R} d \phi
ΔQ=N(Δϕ)R\Delta Q=\frac{N(\Delta \phi)}{R}
ΔQ=Δϕtotal R\Delta Q=\frac{\Delta \phi_{\text {total }}}{R}
=(NBA)R=\frac{(N B A)}{R}
=μ0niπr2R=\frac{\mu_{0} n i \pi r^{2}}{R}
Putting values
=4π×107×100×4×π×(0.01)210π2=\frac{4 \pi \times 10^{-7} \times 100 \times 4 \times \pi \times(0.01)^{2}}{10 \pi^{2}}
ΔQ=32μC\Delta Q=32\, \mu C