Find the magnetic field at point P due to a straight line segment AB of length 6 cm carrying a current of 5A. (See figure) ($$\mu $$₀ = 4$$\pi $$ × 10^–7 N-A^–2)

A thin ring of 10 cm radius carries a uniformly distributed charge. The ring rotates at a constant angular speed of 40 $$\pi $$ rad s^–1 about its axis, perpendicular to its plane. If the magnetic field at its centre is 3.8 × 10^–9 T, then the charge carried by the ring is close to ($$\mu $$₀ = 4$$\pi $$ × 10^–7 N/A² ).

A square loop is carrying a steady current I and the magnitude of its magnetic dipole moment is m. if this square loop is changed to a circular loop and it carries the same current, the magnitude of the magnetic dipole moment of circular loop will be:

The magnitude of the magnetic field at the centre of an equilateral triangular loop of side 1 m which is carrying a current of 10 A is :
[Take $$\mu $$₀ = 4$$\pi $$ × 10^–7 NA^–2]