26. A single-turn wire loop is 2.0 cm in diameter and carries a 650- mA current. Find the magnetic field strength (a) at the loop center and (b) on the loop axis, 20 cm from the center. Wolfson, Richard. Essential University Physics, Volume 2 (p. 511). Pearson Education. Kindle Edition.

(a) Magnetic field at the center of the loop is 4.08 x 10⁻⁵ T

(b) Magnetic field at the axis of the loop is 5.09 x 10⁻⁹ T

Explanation:

Given :

Diameter of the circular loop = 2 cm

Radius of the circular loop, R = 1 cm = 0.01 m

Current flowing through the circular wire, I = 650 mA = 650 x 10⁻³ A

(a) Magnetic field at the center of circular loop is determine by the relation:

Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ T m²/A.

Substitute the suitable values in the above equation.

B = 4.08 x 10⁻⁵ T

(b) Distance from the center of the loop, z = 20 cm = 0.2 m

Magnetic field at the point on the axis of the loop is determine by the relation:

B = 5.09 x 10⁻⁹ T

c, 1a hope this

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