Question

Superconductors can carry very large currents with no resistance. If a superconducting wire is formed into a solenoid of length 20.0 cm with 691 turns, what is the magnetic field inside the solenoid when the current is 5.38 kA? (µ0 = 4π × 10−7 T⋅m/A)

Answer 1

23.36 T

Step-by-step explanation:

We are given that

Length of solenoid=l=20 cm=
`(20)/(100)=0.2 m`

1 m=100 cm

Number of turns=N=691

Current=I=
`5.38kA=5.38* 10^3 A`

`1 kA=10^3 A`

`\mu_0=4\pi* 10^(-7)Tm/A`

We have to find the magnetic field inside the solenoid.

We know that the magnetic field of solenoid

`B=(\mu_0NI)/(l)`

Substitute the values

`B=(4\pi* 10^(-7)* 691* 5.38* 10^3)/(0.2)`

`B=23.36 T`