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)

An electron, moving along the x-axis with an initial energy of 100 eV, enters a region of magnetic field $$\overrightarrow B = \left( {1.5 \times {{10}^{ - 3}}T} \right)\widehat k$$ at S (See figure). The field extends between x = 0 and x = 2 cm. The electron is detected at the point Q on a screen placed 8 cm away from the point S. The distance d between P and Q (on the screen) is : (electron’s charge = 1.6 × 10^–19 C, mass of electron = 9.1 × 10^–31 kg)

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 ).

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]