A wire carrying a current of 26.9 A is bent into a circular arc with a radius of 0.6 cm that sweeps out 0.900 radians. What is t

Question

2 years ago

15

he magnetic field at the center of the arc

1 answer:

melomori [17] · Answer 1 · 2023-02-12T07:59:53+03:00

The magnetic field at the center of the arc is 4 × 10^(-4) T.

To find the answer, we need to know about the magnetic field due to a circular arc.

<h3>What's the mathematical expression of magnetic field at the center of a circular arc?</h3>

According to Biot savert's law, magnetic field at the center of a circular arc is

B=( μ₀I/4π)×(angle/radius)

<h3>What will be the magnetic field at the center of a circular arc, if the arc has current 26.9 A, radius 0.6 cm and angle 0.9 radian?</h3>

B=(μ₀ I/4π)× (0.9/0.006)

= (10^(-7)× 26.9)× (0.9/0.006)

= 4 × 10^(-4) T

Thus, we can conclude that the magnitude of magnetic field at the center of the circular arc is 4 × 10^(-4) T.

Learn more about the magnetic field of a circular arc here:

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