Question

A 100 A current circulates around a 1.60-mm-diameter superconducting ring. What is the on-axis magnetic field strength 5.40 cm from the ring? Express your answer with the appropriate units.

Answer 1

magnetic field is 2.55 ×
`10^(-7)` T

Step-by-step explanation:

given data

current = 100 A

diameter = 1.60 mm

strength = 5.40 cm

to find out

What is the on-axis magnetic field

solution

we know here magnetic field on x axis current carry circular loop so it will

B = u / 4π × 2π R² I /
`( x^(2) + R^(2) )^(3/2)` ...............1

here I is current and r is radius and x distance

so

B = 4π ×
`10^(-7)` / 4π × 2π (1.60× 10^-3 / 2 )² (100) /
`( (5.40*10^(-2))^(2) + (1.60*10^(-3)/ 2 )^(2) )^(3/2)`

solve it and we get B

B = 2.55 ×
`10^(-7)` T

magnetic field is 2.55 ×
`10^(-7)` T