two protons P and Q moving with a same speed enters magnetic field B1 and B2 respectively at right angles to the field direction. if B1>B2, for which proton the circular path in the magnetic field have smaller radius. Send me today I need it very urgently please.

Dear Student,

Force on a charge particle due to magnetic field= F= q( v x B) 
now Magnetic field is directed perpendicular to motion, sin(@) =1  or F =qvB 

Now this particle under this force will start moving in a circle, or in other words experience centripetal force.

F= mv²/R   where R=radius of the circle.

Now this centrifugal force = magnetic force for a circular motion.

or $$\begin{gathered} \frac{m{v}^2}{R}=qvB \\ or R=\frac{mv}{qB} \\ given other terms were same/cons\tan t. \\ R \propto \frac{1}{B} \\ or, \frac{R_1}{R_2}=\frac{B_2}{B_1} \\ Now \sin ce R is inversely proportional to B so if B is larger R will be lesser. \\ Thus as B_1>B_{2 } we have R_1<R_2 \\ i.e. proton P will have a lesser radius. \end{gathered}$$

Regards.