Question

A solenoid passing by a current of 5.0 A generates a magnetic field at its diameter of 50 μT. Thus the number of spirals per length scale is:

A. 5.0 / π Spear / m

B. 10 / π Spear / m

C. 20 / π Spear / m

D. 25 / π Spear / m

Answer 1

D. 25 / π Spiral / m

Step-by-step explanation:

Given;

current, I = 5 A

magnetic field strength, B = 50 μT = 50 x 10⁻⁶ T

The magnetic field strength is given as;

`B = \mu_0 nI\\\\where;\\\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi * 10^(-7) T/A.m\\\\n \ is \ the \ number \ of \ spirals \ per \ length\\\\n = (B)/(\mu_0 I) = (50 * 10^(-6))/(5* 4\pi * 10^(-7)) = (25)/(\pi ) \ spirals /m \\\\`

Therefore, the correct option is D. 25 / π Spiral / m