Question

A cylindrical solenoid 34 cm long with a radius of 7 mm has 475 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 130 ohms. One microsecond after connecting the loose wire to the battery to form a series circuit with the battery and a 20 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.) counter-clockwise O clockwise 2092

Answer 1

current through the solenoid,

I = V/R

= 9/20

= 0.45 A

magnetic field inside the solenoid,

B = mue*N*I/L

induced emf in the loop = N'*A*dB/dt

= N'*pi*r^2*(mue*N/L)*(dI/dt)

= 2*pi*0.007^2*(4*pi*10^-7*475/0.34)*(0.45/10^-6)

= 0.2432 V

induced current in the loop = induced emf/R

= 0.2432/130

= 1.87*10^-3 A

direction : clockwise