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

35.7 mA

Step-by-step 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=(N)/(L)=(870)/(0.390)=2230.8`

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

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