In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt. \(\frac{E^{a}}{2 b} \mathrm{~J} / \mathrm{s}\) represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of \(\frac{b}{a}\) will be ______

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Electromagnetic Induction

In the given figure, an inductor and a resistor are connected in series with a battery of emf E volt.

\frac{E ^{a}}{2 b} J / s

represents the maximum rate at which the energy is stored in the magnetic field (inductor). The numerical value of

\frac{b}{a}

will be ______

Answer: 25

The energy stored in the inductor is

U = \frac{1}{2} L I^{2}

, so its rate of storage is

\frac{d U}{d t} = L I \frac{d I}{d t} .

During current growth in the

R

L

circuit,

I = \frac{E}{R} (1 - e^{- R t / L})

and

\frac{d I}{d t} = \frac{E}{L} e^{- R t / L}

, hence

\frac{d U}{d t} = \frac{E ^{2}}{R} (1 - e^{- R t / L}) e^{- R t / L} .

Putting

u = e^{- R t / L}

, the factor

u (1 - u)

is greatest at

u = \frac{1}{2}

, where it equals

\frac{1}{4}

, so

(\frac{d U}{d t})_{m a x} = \frac{E ^{2}}{4 R} .

With

R = 25 Ω

from the circuit,

(\frac{d U}{d t})_{m a x} = \frac{E ^{2}}{100} = \frac{E ^{2}}{2 \times 50} .

Comparing with

\frac{E ^{a}}{2 b}

gives

a = 2

and

b = 50

, so

\frac{b}{a} = 25

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer