Question

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.

Answer 1

Answer:

(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:

$B=\frac{\mu_{0}I }{2R}$

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.

$B=\frac{4\pi\times10^{-7}\times650\times10^{-3} }{2\times0.01}$

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:

$B=\frac{\mu_{0}IR^{2} }{2(z^{2}+R^{2})^{3/2} }$

$B=\frac{4\pi\times10^{-7}\times650\times10^{-3}\times (0.01)^{2} }{2((0.2)^{2}+(0.01)^{2})^{3/2} }$

B = 5.09 x 10⁻⁹ T