Answer: The magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.
Explanation:
Given: Length = 0.50 m
No. of turns = 500
Current = 22 A
Formula used to calculate magnetic field is as follows.
[tex]B = \mu_{o}(\frac{N}{L})I[/tex]
where,
B = magnetic field
[tex]\mu_{o}[/tex] = permeability constant = [tex]4\pi \times 10^{-7} Tm/A[/tex]
N = no. of turns
L = length
I = current
Substitute the values into above formula as follows.
[tex]B = \mu_{o}(\frac{N}{L})I\\= 4 \pi \times 10^{-7} Tm/A \times (\frac{500}{0.5 m}) \times 22\\= 0.0276 T[/tex]
Thus, we can conclude that magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.