Question #a8f63

1 Answer
Sep 11, 2015

Answer:

#=1,258 mum#

Explanation:

Note that the unit of magnetic flux density (B) is the Tesla, which is dimensionally equivalent to a #(Wb)/m^2#, since by definition it is the magnetic flux per unit area. (incorrect SI unit given in the original question).

The velocity v of the proton may be found from the definition of potential difference, together with the principle of conservation of energy. Since by definition, potential difference between 2 points is defined as the work done per unit charge to move a positive charge between the 2 points, and this work done is transferred into kinetic energy of the proton, it implies that
#QV=1/2 mv^2 => v=sqrt((2QV)/m)#
#=sqrt(((2)(9,1xx10^(-31))(6xx10^5))/(1,67xx10^(-27))) = 25,571 m//s #

Now, an electric charge Q moving at velocity v in the presence of a magnetic field of flux density B, experiences a magnetic force given by
#vecF=QvecvxxvecB#

This is a vector cross product so its direction can be obtained from the right hand rule, and is perpendicular to both v and B.
This force is hence directed towards the centre all the time and is thus the required centripetal force #F_c# to keep the charge moving in a circular motion.

The magnitude of this force can be given by #Q*||vecv||*||vecB ||*sintheta# , where #theta# is the angle between #vecv and vecB#

So, in this case, #QvBsintheta=(mv^2)/r#
#therefore r=(mv)/(QBsintheta)#
#=((1,67xx10^(-27))(25,571))/((1,6xx10^-19)(0,3)sin45^@)#

#=1,258 mum#