Question

A long solenoid that has 920 turns uniformly distributed over a length of 0.380 m produces a magnetic field of magnitude 1.00 10-4 T at its center. What current is required in the windings for that to occur?

Answer 1

Answer: 0.0328 A

Explanation:

Given

No of turns $N=920$

Length of solenoid $L=0.380\ m$

Magnetic field $B=10^{-4}\ T$

the magnetic field at the center of the solenoid is

$\Rightarrow B=\mu nI=\mu \dfrac{N}{L}I$

Putting values we get

$\Rightarrow 10^{-4}=4\pi \times 10^{-7}\times \dfrac{920}{0.380}\times I\\\\\Rightarrow I=\dfrac{0.380\times 1000}{920\times 4\pi}=0.0328\ A$