A wire A, bent in the shape of an arc of a circle, carrying a current of 2 A and having radius 2 cm and another wire B, also bent in the shape of arc of a circle, carrying a current of 3 A and having radius of 4 cm, are placed as shown in the figure. The ratio of the magnetic fields due to the wires A and B at the common centre O is :

A charged particle carrying charge 1 $$\mu $$C is moving
with velocity $$\left( {2\widehat i + 3\widehat j + 4\widehat k} \right)$$ ms^–1. If an external
magnetic field of $$\left( {5\widehat i + 3\widehat j - 6\widehat k} \right)$$× 10^–3 T exists in the region where the particle is moving then the
force on the particle is $$\overrightarrow F $$ × 10^–9 N. The vector $$\overrightarrow F $$ is :

Magnitude of magnetic field (in SI units) at the centre of a hexagonal shape coil of side 10 cm, 50 turns and carrying current I (Ampere) in units of $${{{\mu _0}I} \over \pi }$$ is :

The figure shows a region of length ‘l’ with a uniform magnetic field of 0.3 T in it and a proton entering the region with velocity 4 $$ \times $$ 10⁵ ms^–1 making an angle 60^o with the field. If the proton completes 10 revolution by the time it cross the region shown, ‘l’ is close to
(mass of proton = 1.67 $$ \times $$ 10^–27 kg, charge of the proton = 1.6 $$ \times $$ 10^–19 C)