Connect a battery to a solenoid A cylindrical solenoid 42 cm

Connect a battery to a solenoid A cylindrical solenoid 42 cm long with a radius of 8 mm has 200 tightly-wound turns of wire uniformly distributed along its length (see the figure). Around the middle of the solenoid is a two-turn rectangular loop 3 cm by 2 cm made of resistive wire having a resistance of 175 ohms. One microsecond after connecting the loose wire to the battery to form a series circuit with the battery and a 20 Q resistor, what is the magnitude of the current in the rectangular loop and its direction (clockwise or counter-clockwise in the diagram)? (The battery has an emf of 9 V.)

Solution

N1 = 200

l = 42 cm = 0.42 m

r = 8 mm = 0.008 m

R1 = 175 ohm

N2 = 2

R2 = 2 ohm

Inductance of the solenoid, L = u*N1*I = u*N1*Io*(1 - e^(-t*R2/L))

where Io = initial current through the solenoid, = V/R2 = 9/20 = 0.45 A

So, L = 1.26*10^-6*200*0.45*(1 - e^(-t*2/L))

At t = 1 us ,

L = 1.26*10^-6*200*0.45*(1 - e^(-(1*10^-6)*2/L))

Soo, L = 1.46*10^-5 H

So, Current in the inductor = 0.45*(1 - e^(-(1*10^-6)*2/(1.46*10-5)))

= 0.058 A

So, flux in the inductor in time t = N2 *L*I =  2*1.26*10^-6*200*(0.45*(1 - e^(-(1*10^-6)*2/L)))^2

So, emf in time t = 1 us,

E =  2*1.26*10^-6*200*2*(0.45*(1 - e^(-(1*10^-6)*2/L)))*(0.45*(2/L)*e^(-t*2/L))

= 2*1.26*10^-6*200*2*(0.45*(1 - e^(-(1*10^-6)*2/(1.46*10^-5))))*(0.45*(2/(1.46*10^-5))*e^(-(1*10^-6)*2/(1.46*10^-5)))

= 3.12 V

So, current in the rectangular loop = 3.12/175 = 0.0178 A