Two circular coils having radii of 5.0 cm and 10 cm carry equal currents of 2.0 A . The coils consist of 50 and 100 turns respectively, and they are positioned so that their centres and planes coincide. If the outer coil is subsequently rotated by 90^ about a diameter, determine the magnitude of the magnetic field B at the centre.

Step-by-step solution

The magnetic field at the centre of a circular coil of N turns, radius R , carrying current I is given by B = ₀ N I 2 R . For the inner coil: B₁ = 4 10⁻⁷ 50 2.0 2 5.0 10⁻² = 4 10⁻⁴ T For the outer coil: B₂ = 4 10⁻⁷ 100 2.0 2 10 10⁻² = 4 10⁻⁴ T When the outer coil is rotated by 90^ about its diameter, its plane becomes perpendicular to the plane of the inner coil. Consequently, their magnetic fields at the centre will be mutually perpendicular. The resultant magnetic field B is: B = B₁^2 + B₂^2 B = (4 10⁻⁴)^2 + (4 1