Question

A solenoid is built with 870 turns uniformly distributed over a length of 0.390 m to produce a magnetic field of magnitude 1.00 × 10−4 T at its center. What is current necessary for this to happen? 17.4 mA

Answer 1

Answer:

35.7 mA

Explanation:

The magnetic field inside a solenoid is given by:

$B=\mu_0 I n$ (1)

where

$\mu_0 = 4\pi \cdot 10^{-7} H/m$ is the vacuum permeability

I is the current

n is the number of turns per unit length

Since we have

N = 870 turns

L = 0.390 (length of the solenoid)

we can calculate n

$n=\frac{N}{L}=\frac{870}{0.390}=2230.8$

And now we can re-arrange eq.(1) to find the current, I:

$I=\frac{B}{\mu_0 n}=\frac{1.00\cdot 10^{-4} T}{(4\pi\cdot 10^{-7} H/m)(2230.8m^{-1})}=0.0357 A = 35.7 mA$