Question

A 500-turn solenoid, 25 cm long, has a diameter of 2.5 cm. A 10-turn coil is wound tightly around the center of the solenoid. If the current in the solenoid increases uniformly from 0 to 5.0 A in 0.60 s, what will be the induced emf in the short coil during this time?

Answer 1

`E=1.028x10^(-4) V`

Step-by-step explanation:

`N=500\\n=10`

`d=25cm*(1m)/(100cm)=0.25m`

`I=5A\\t=0.6s`

`u_(o)=4\pi x10^(-7)(T*m)/(A)`

So the field create inside the coil can be find:

`\beta =(u_(o)*N*i)/(L)`

The rate of tha change of B field is:

Δβ/Δt=
`(u_(o)*N*(di)/(dt))/(L)`

`(di)/(dt)=(5A)/(0.6s)=8.333 (A)/(s)`

Δβ/Δt=
`(4\pi x10^(-7)(T*m)/(A) *500*8.33(A)/(s))/(0.25m)`

Δβ/Δt=
`20.935x10^(-3) (T)/(s)`

Induce E caused by tha changing B is:

E=[(Δβ/Δt)*A*n]

`r=2.5cm=r=0.025m(0.025m)/(2)=0.0125m\\A=\pi *r^(2)=\pi*(1.25x10^(-2)m)^(2)\\A=4.91x10^(-4) m^(2)\\E=(2.09x10^(-2)(T)/(s)*4.91x10^(-4)m^(2)*10)  \\`

E=1.028
`x10^(-4)`V