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 above the centre of the loop.

Step-by-step solution

The magnetic field at a distance x on the axis of a circular loop of radius R carrying current I is given by: B = ₀ I R^2 2(R^2 + x^2)^ 3/2 Given values are: I = 5.0 A R = 4.0 cm = 4.0 10⁻² m x = 3.0 cm = 3.0 10⁻² m ₀ = 4 10⁻⁷ T m/A Calculating the denominator term: R^2 + x^2 = (4.0 10⁻²)^2 + (3.0 10⁻²)^2 = 25 10⁻⁴ m ^2 (R^2 + x^2)^ 3/2 = (25 10⁻⁴)^ 3/2 = 125 10⁻⁶ m ^3 Substituting the values into the formula: B = 4 10⁻⁷ 5.0 (16 10⁻⁴) 2 125 10⁻⁶ B = 320 10⁻¹¹ 250 10⁻⁶ B = 1.28 10⁻⁵ T Using 3.14 : B 4.0 10⁻⁵ T Since