A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the magnetic fields due to the wire at distance $${a \over 3}$$ and 2$$a$$, respectively from the axis of the wire is :

A charged particle of mass 'm' and charge 'q' moving under the influence of uniform electric field $$E\overrightarrow i $$ and a uniform magnetic field $$B\overrightarrow k $$ follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, $$v\overrightarrow i $$ and $$ - 2v\overrightarrow j $$ . Then which of the following statements (A, B, C, D) are the correct ?
(Trajectory shown is schematic and not to scale) :
(A) E = $${3 \over 4}\left( {{{m{v^2}} \over {qa}}} \right)$$

(B) Rate of work done by the electric field at P is $${3 \over 4}\left( {{{m{v^3}} \over a}} \right)$$

(C) Rate of work done by both the fields at Q is zero

(D) The difference between the magnitude of angular momentum of the particle at P and Q is 2mav.

A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is :

Photon with kinetic energy of 1MeV moves from south to north. It gets an acceleration of 10¹² m/s² by an applied magnetic field (west to east). The value of magnetic field : (Rest mass of proton is 1.6 × 10^–27 kg) :