Question

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?

Answer 1

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