Maybe this will help...
https://www.universetoday.com/tag/isaac-newton/
Answer:
I believe the answer is D
Answer:
0.957mm
Explanation:
The on-axis magnetic field of a current loop(B_loop) is is 4.9 nT
B_loop = (2μAI) / (4πz^3).
Where z is defied as distance from the current carrying loop,
A is the total enclosed area of the loop,
I is the current,
μ is the permeability constant (4π*10^-7).
4.9 nT = (2* 4π*10^-7 *A *32A) / (4π* (50cm)^3)
4.9 *10^-9 = 5.12*10^-5 A
A = 0.0000957 m
=0.957mm
Go into place with no windows and duck and cover
