A battery of emf e is connected with a small circular wire loop of radius r and resistance R. Suddenly due to some reason battery get short circuited internally at t = 0. For t > 0 (consider uniform magnetic field in the loop)

Engineering

Physics

LR Circuit Growth and Decay of Current

Self and Mutual Inductance

Magnetic Flux and Faraday Law and Lenz Law

Question

A battery of emf e is connected with a small circular wire loop of radius r and resistance R. Suddenly due to some reason battery get short circuited internally at t = 0. For t > 0 (consider uniform magnetic field in the loop)

the current in the loop as a function of time is given by $\frac{ε}{R} e^{- 2 Rt / μ_{0} πr}$

the current in the loop become zero.

The time in which current in loop decreases to 10^–3 $\frac{ε}{R}$ (from t = 0) for R = 100 $Ω$ and r = 5 cm is given by $\frac{3 π^{2}}{10^{6}} s$

The time in which current in loop decreases to 10^–3 $\frac{ε}{R}$ (from t = 0) for R = 100 $Ω$ and r = 5 cm is given by $\frac{3 π^{2} \ln 10}{10^{10}} s$

LiveFree JEE Main Previous Year Online Papers Live Free JEE Advance Previous Year Online Papers

College PredictorLive

Know your College Admission Chances Based on your Rank/Percentile, Category and Home State.

Get your JEE Main Personalised Report with Top Predicted Colleges in JoSA

Flux linked with loop due to its own magnetic field,

$ϕ = \frac{μ_{0} I}{2 r} ({πr}^{2}) = \frac{μ_{0} πrI}{2}$

emf induced = – $\frac{dϕ}{dt} = ε = \frac{μ_{0} πr}{2} \frac{dI}{dt}$

$I = \frac{ε}{R} = \frac{μ_{0} πr}{2 R} \frac{dI}{dt}$

$\int_{I_{0}}^{I} \frac{dI}{I} = - \int_{0}^{t} \frac{2 R}{μ_{0} πr} \cdot dt = I_{0}^{} e^{- 2 Rt / μ_{0} πr}$

Now, 10^–3 $I_{0} = I_{0} e^{- \frac{2 Rt}{μ_{0} πr}}$

which gives $t = \frac{3 π^{2} \ln 10}{10^{10}} s$