A small square loop carries the current 32A. the on axis magnetic field strength 50 cm for the loop is 4.9 nT. What is the edge length of the square?

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

answer; i believe that the correct answer is (tabula rasa);

the answer is a wheel