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3 years ago

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A 100 A current circulates around a 1.60-mm-diameter superconducting ring. What is the on-axis magnetic field strength 5.40 cm f

rom the ring? Express your answer with the appropriate units.

1 answer:

Aleksandr [31]3 years ago

6 0

Answer:

magnetic field is 2.55 × $10^{-7}$ T

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

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Answer:

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Explanation:

We are given that

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Answer:

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