Situated in a horizontal plane, a circular loop of radius 4.0 cm carries an electric current of 5.0 A in the clockwise direction when viewed from above. Determine the magnetic field at a point 3.0 cm below the centre of the loop.

Step-by-step solution

The magnitude of the magnetic field on the axis of a circular loop is given by: B = ₀ I R^2 2(R^2 + z^2)^ 3/2 Given values are: I = 5.0 A R = 4.0 cm = 0.04 m z = 3.0 cm = 0.03 m ₀ = 4 10⁻⁷ T m/A Substituting the values into the formula: B = 4 10⁻⁷ 5.0 (0.04)^2 2((0.04)^2 + (0.03)^2)^ 3/2 B = 4 10⁻⁷ 5.0 1.6 10⁻³ 2(1.6 10⁻³ + 0.9 10⁻³)^ 3/2 B = 32 10⁻¹⁰ 2(2.5 10⁻³)^ 3/2 B = 32 10⁻¹⁰ 2(125 10⁻⁶) B = 32 10⁻¹⁰ 250 10⁻⁶ = 1.28 10⁻⁵ T Using 3.14 : B 1.28 3.14 10⁻⁵ T 4.0 10⁻⁵ T According to the right-hand thumb rule, since