A thin 12 cm long solenoid has a total of 460 turns of wire and carries a current of 2.4 A?

Question

Calculate the magnitude of the field inside near the center. T


I would be VERY greatful for help w/ this!

Answer 1

Ok, here we go :

You need to remeber, that the electromagnetic field near the center of a solenoid is :

B = Uo*I*(N/L)

Where :

Uo = 4*pi*10^-7

I = current

N = 460 turns

L = 12*10^-2 meters

now, replacing :

B = 4*pi*10^-7*2.4*460 / 12*^10-2

pi = 3.1416, sorry I don't know the symbol here.

B = 368*pi*10^-5 = 0.01 (Tesla)

Hope that might help you.

Answer 2

Magnitude of magnetic field at a point on the axis and near the center of a solenoid=4pix10^ -7 x number of turns per metrexcurrent=4 pix10 ^ -7x460x2.4/0.12 tesla=1.11x10^-3 tesla