Question

A long solenoid that has 1,030 turns uniformly distributed over a length of 0.410 m produces a magnetic field of magnitude 1.00 multiply.gif 10-4 T at its center. What current is required in the windings for that to occur? mA

Answer 1

Given

magnetic field B = $1*10^-4 T$

total number of turn N = 1030

total length L = 0.410 m

number o turn per unit length   $n = \frac{N}{L}$

let current is i then magnetic field   $B = \mu _{0} ni$

$1*10^-4 = \mu _{0} \frac{1030}{0.410}*i$

$i = \frac{1*10^-4 *0.410}{\mu _{0 * 1030}}$

$i = 0.03167 A = 31.67 m A$

answer is 31.67 mA

Thanks for asking.