Question

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 the magnetic field at the center of the arc

Answer 1

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.

What's the mathematical expression of magnetic field at the center of a circular arc?

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

• B=(μ₀ I/4π)× (arc/radius²)

• As arc is given as angle × radius, so

        B=( μ₀I/4π)×(angle/radius)

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?

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:

brainly.com/question/15259752

#SPJ4