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

Answer:

23.36 T

Explanation:

We are given that

Length of solenoid=l=20 cm=$\frac{20}{100}=0.2 m$

1 m=100 cm

Number of turns=N=691

Current=I=$5.38kA=5.38\times 10^3 A$

$1 kA=10^3 A$

$\mu_0=4\pi\times 10^{-7}Tm/A$

We have to find the magnetic field inside the solenoid.

We know that the magnetic field of solenoid

$B=\frac{\mu_0NI}{l}$

Substitute the values

$B=\frac{4\pi\times 10^{-7}\times 691\times 5.38\times 10^3}{0.2}$

$B=23.36 T$