An ideal solenoid is 18.5 cm long, has a circular cross-section 2.20 cm in diame

Question

An ideal solenoid is 18.5 cm long, has a circular cross-section 2.20 cm in diameter, and contains 545 equally spaced thin windings. This solenoid is connected in a series circuit with a 15 ohm resistor, a battery of internal resistance ohms and open-circuit terminal voltage of 25 V, and an open switch. (Note u = 4*(pi)*10-7 T *m/A) How long after closing the switch will it take for the stored energy in the solenoid to reach 1/2 of its maximum value?

Reference https://www.physicsforums.com/threads/current-decay-in-an-r-l-circuit.420346/

Explanation / Answer

We know -
L = (uoN2A)/l
where L = inductance in Henry
N = number of turns
A = area of cross-section
l = length in meters
Substituting Values -

L = (4 * 3.14 * 10^-7 * 545^2 * 3.14 * (2.2*10^-2 )^2) / 0.185
L = 7.66 *10-4 H

R = 20 ohm

Maximum Energy stored in an inductor = 1/2 L Imax2
This happens when the current is max.

Let value of current, when stored energy of solenoid is 1/2 of it's max Value = I

Solving for I

1/2 L * I2 = 1/2 * (1/2 * L * Imax2)
I = Imax/sqrt(2)

Time required for current to reach above value is the time taken for the energy in solenoid to reach 1/2 of it's max value.

Now, time t needed for current to reach this value is given by -

I = Imax (1 - e-t/tau)
Substituting Value of tau and taking log of each side , we solve for t

t = -L/R ln(1 - I/Imax)
t = - 7.66 *10-4 / 20 ln(1- Imax/sqrt(2) /Imax)
t = - 3.83 * 10^-5 ln (1 - 0.707)
t = - 3.83 * 10^-5 * -1.22
t = 4.70 * 10^-5 Seconds

Time needed for the stored energy in solenoid to reach 1/2 of it's maximum value = t = 4.70 * 10^-5 Seconds

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