Question

a) Assume that there is a long ideal solenoid with 120 turns/cm. It carries a current i= 2 A and it has a diameter 2 cm and a length 5 m. Find the uniform magnetic field inside the solenoid.

c) Now, construct an RL circuit using an ideal battery that has potential difference 5 V, one resistor with R = 2 Ω and the solenoid that has same shape with one mentioned at part (a). Wait very long time and then remove the battery. Find the potential difference across resistor at 2 ms after removing battery? (Hint: First calculate the inductance)

Answer 1

a)Magnetic field in the solenoid B=nI

=4x

n=number of turns per unit length=120/cm=12000/m

I=current=2A

B=nI=4xx12000x2=3.01xT

c)Current at t=2ms

I=

R=resistance=2

=maximum current =V/R=5/2=2.5A

t=2ms=0.002s

L=inductance of solenoid =NnA

A=area of cross section==3.14x =3.14xm2 (r=diameter/2=2/2=1cm=0.01m)

N=total number of turns=nxlength=12000x5=60000

L=NnA=4xx60000x12000x3.14x=0.284H

I==2.5=2.5x1.014=0.986A