Question

The magnetic field at a point inside a 2.0 mH inductor-coil becomes 0.80 of its maximum value in 20 µs when the inductor is joined to a battery. Find the resistance of the circuit.

Answer 1

Given:-

Inductance of the inductor, L = 2.0 mH

Let the resistance in the circuit be R and the steady state value of the current be i₀.

At time t , current i in the LR circuit is given by

i = i₀(1 − e^−t/τ)

Here,

\[\tau = \frac{L}{R}=\] Time constant

On multiplying both sides by µ₀n, we get

n = Number of turns per unit length of the coil

µ₀ni = µ₀ni₀(1 − e^−t/τ)

⇒ B = B₀(1 − e^−tR/L)

⇒ 0.8 B₀ = B₀

\[\left( 1 - e^\frac{- 20 \times {10}^{- 6} R}{2 \times {10}^{- 3}} \right)\]

⇒ 0.8 = (1 − e^−R/100)

⇒ e^−R/100 = 0.2

⇒ ln (e^−R/100) = ln (0.2)

`rArr -R/100=-1.693`

⇒ R = 169.3  Ω